# ML Aggarwal Solutions for Chapter 5 Quadratic Equations in One Variable Class 10 Maths ICSE

Here, we are providing the solutions for Chapter 4 Linear Inequation from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of the fourth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 4 Linear Inequation ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on the equations of algebra, word problems, representation on number line and solutions of set.

### Exercise 5.1

1. Check whether the following are quadratic equations:

(i) x2 – 2x + 3/5 = 0

(ii) (2x + 1) (3x – 2) = 6(x + 1)(x – 2)

(iii) (x – 3)3 + 5 = x3 + 7x2 – 1

(iv) x – 3/x = 2, x 0

(v) x + 2/x = x2, x ≠ 0

(vi) x2 + 1/x2 = 3, x ≠ 0

(i) √3x2 – 2x + 3/5 = 0

It is a quadratic equation as it is power of 2.

(ii) (2x + 1)(3x – 2) = 6(x + 1)(x – 2)

6x2 – 4x + 3x – 2 = 6(x2 – 2x + x – 2)

6x2 – x – 2 = 6x2 – 12x + 6x – 12

12x – 6x – x = - 12 + 2

5x = - 10

It is not a quadratic equation.

(iii) (x – 3)3 + 5 = x3 + 7x2 – 1

x3 – 3x2 × 3 + 3x × 9 – 27 + 5

= x3 + 7x2 – 1

- 9x2 + 27x – 22 – 7x2 + 1 = 0

- 16x2 + 27x – 21 = 0

⇒ 16x2 – 27x + 21 = 0

(iv) x – 3/x = 2, x ≠ 0

x2 – 3 = 2x ⇒ x2 – 2x – 3 = 0

(v) x + 2/x = x2, x ≠ 0

x2 – 3 = x3

⇒ x3 – x2 – 2 = 0

It is not a quadratic equation.

(vi) x2 + 1/x2 = 3, x ≠ 0

x4 + 6 = 3x2

x4 – 3x2 + 6 = 0

It is not a quadratic equation.

2. In each of the following, determine whether the given numbers are roots of the given equations or not;

(i) x2 – x + 1 = 0 ; 1, - 1

(ii) x2 – 5x + 6, 0; 2, - 3

(iii) 3x2 – 13x – 10 = 0, 5, -2/3

(iv) 6x2 – x – 2 = 0; -1/2, 2/3

(i) x2 – x + 1 = 0; 1, - 1

Where x = 1, then

(1)2 – 1 + 1 = 1 – 1 + 1 = 1 ≠ 0

∴ x = 1 does not satisfy it

And (-1)2 – (-1) + 1 = 0

1 + 1 + 1 ⇒ 3 ≠ 0

∴ x = - 1, does not satisfy it

∴ x = 1, - 1 are not roots of the equation.

(ii) x2 – 5x + 6 = 0; 2, - 3

When x = 2, then

(2)2 – 5 × 2 + 6 = 4 – 10 + 6 = 10 – 10 = 0

∴ x = 2 is its root.

When, x = - 3, then

(-3)2 – 5(- 3) + 6

= 9 + 15 + 6

= 30 ≠ 0

∴ x = - 3 is not is solution

∴ 2 is root of the equation by – 3 is not a root.

(iii) 3x2 – 13x – 10 = 0; 5, -2/3

x = 5,

3(5)2 – 13 × 5 – 10 = 75 – 65 – 10

= 75 – 75

= 0

∴ x = 5 is its root

If x = - 2/3, then

3(-2/3)2 – 13 × -2/3 – 10

= (3 × 4)/9 + 26/3 – 10

= 4/3 + 26/2 – 10

= 30/3 – 10

= 10 – 10

= 0

∴ x = -2/3 is also its root.

Hence, both 5, -2/3 are its roots.

(iv) 6x2 – x – 2 = 0; -1/2, 2/3

If x = -1/2, then

= 6(-1/2)2 – (-1/2) – 2

= 6 × 1/4 + 1/2 – 2

= 3/2 + 1/2 – 2

= 4/2 – 2

∴ x = -1/2 is its root

If x = 2/3, then

= 6 × 4/9 – 2/3 – 2

= 8/3 – 2/3 – 2

= 6/3 – 2

= 0

∴ x = 2/3 is also its root.

Hence -1/2, 2/3 are both its root.

3. In each of the following, determine whether the given numbers are solutions of the given equation or not:

(i) x3 – 3√3x + 6 = 0; √3, - 2√3

(ii) x2 = √2x – 4 = 0, x = - √2, 2√2

(i) x2 - 3√3x + 6 = 0; √3, -2√3

(a) Substituting the value of x = 3

L.H.S. = x2 - 3√3x + 6

= (√3)2 - 3√3 ×√3 + 6

= 3 – 9 + 6 = 0

= R.H.S.

∴ x = 3 is its solution

(b) x = -2√3

Substituting x = -23

L.H.S. = x2 - 3√3x + 6

= (-2√3)2 - 3√3 (-2√3) + 6

= 12 + 18 + 6

= 36 ≠ 0

∵ x = -2√3 is not its solution

(ii) x2 - √2x – 4 = 0, x = -√2, 2√2

(a) x = - √2

Substituting x = - √2

L.H.S. = x2 - √2x – 4

= (-2)2 - √2(-√2) – 4

= 2 + 2 – 4

= 0

= R.H.S.

∴ x = - √2 is its solution

(b) x = -2√2

Substituting = -2√2

L.H.S. = x2 - √2x – 4

= (-2√2)2 - √2 (-2√2) – 4

= 8 – 4 – 4

= 8 – 8

= 0

= R.H.S

∴ x = -2√2 is its solution.

4. (i) If –(1/2) is a solution of the equation 3x2 + 2kx – 3 = 0, find the value of k.

(ii) If 2/3 is a solution of the equation 7x2 + kx – 3 = 0, find the value of k.

(i) x = - 1/2 is a solution of the

3x2 + 2kx – 3 -= 0,

Substituting the value of x in the given equation

3(-1/2)2 + 2k(-1/2) – 3 = 0

3 × 1/4 – k – 3 = 0

3/4 – k – 3 = 0

⇒ k = 3/4 – 3 = - 9/4

Hence k = - 9/4

(ii) 7x2 + kx – 3 = 0, x = 2/3

∵ x = 2/3 is its solution

∴ 7(2/3)2 + k(2/3) – 3 = 0

⇒ 7 × 4/9 + 2/3k – 3 = 0

⇒ 28/9 – 3 + 2/3k = 0

⇒ 2/3k = 3 – 28/9

⇒ 2/3k = (27 – 28)/9

⇒ 2/3k = -1/9

⇒ k = -1/9 × 3/2

= -1/6

Hence k = -1/6

5. (i) If √2 is a root of the equation kx2 + √2 – 4 = 0, find the value of k.

(ii) If a is a root of the equation x2 – (a + b)x + k = 0, find the value of k.

(i) kx2 + √2 – 4 = 0, x = √2

x = √2 is its solution

∴ k(√2)2 + √2×√2 – 4 = 0

⇒ 2k + 2 – 4 = 0

⇒ 2k – 2 = 0

⇒ 2k = 2

⇒ k = 2/2 = 1

∴ k = 1

(ii) x2 – x(a + b) + k = 0, x = a

∵ x = a is its solution

∴ (a)2 – a(a + b) + k = 0

⇒ a2 – a2 – ab + k = 0

⇒ -ab + k = 0

∴ k = ab

6. if 2/3 and -3 are the roots of the equation px2 + 7x + q = 0, find the values of p and q.

2/3 and -3 are the roots of the equation px2 + 7x + q = 0

Substituting the value of x = 2/3 and – 3 respectively, we get

p(2/3)2 + 7(2/3) + q = 0

⇒ 4/9.p + 14/3 + q = 0

⇒ 4p + 9q = - 42 ….(i)

And p(-3)2 + 7(-3) + q = 0

9p – 21 + q = 0

⇒ 9p + q = 21 ....(ii)

Q = 21 – 9p

Substituting the value of q in (i)

4p + 9(21 – 9p) = - 42

4p – 189 – 81p = - 42

-77p = - 42 – 189

= - 231

p = - 231/-77 = 3

∴ q = 21 – 9×3

= 21 – 27

= - 6

∴ p = 3, q = - 6

### Exercise 5.2

1. Solve the following equations (1 to 24) by factorization

(i) 4x2 = 3x

(ii) (x2 – 5x)/2 = 0

(i) 4x2 = 3x

x(4x – 3) = 0

either x = 0,

or 4x – 3 = 0, then 4x = 3

⇒ x = 3/4

∴ x = 0, 3/4

(ii) (x2 – 5x)/2 = 0

x2 – 5x = 0

⇒ x(x – 5) = 0

Either x = 0 or x – 5 = 0, then x = 5

Hence x = 0, 5

2. (i) (x – 3)(2x + 5) = 0

(ii) x(2x + 1) = 6

(i) (x – 3)(2x + 5) = 0

Either x – 3 = 0

Then x = 3

Or 2x + 5 = 0 then 2x = - 5

⇒ x = -5/2

Hence x= 3, = -5/2

(ii) x(2x + 1) = 6

x(2x + 1) = 6

2x2 + x – 6 = 0

2x2 + 4x - 3x – 6 = 0

⇒ 2x(x + 2) – 3(x + 2) = 0

⇒ (x + 2)(2x – 3) = 0

Either x + 2 = 0, then x = - 2

Or 2x – 3 = 0, then 2x = 3

⇒ x = 3/2

Hence x = - 2, 3/2

3. (i) x2 – 3x – 10 = 0

(ii) x(2x + 5) = 3

(i) x2 – 3x – 10 = 0

⇒ x2 – 5x + 2x – 10 = 0

⇒ x(x – 5) + 2(x – 5) = 0

⇒ x(x – 5)(x + 2) = 0

Either x – 5 = 0, then x = 5

Or x + 2 = 0, then x = - 2

Hence, x = 5, - 2

(ii) x(2x + 5) = 3

x(2x + 5) = 3

⇒ 2x2 + 5x + 3 = 0

⇒ 2x2 + 6x – x – 3 = 0

⇒ 2x(x + 3) – 1(x + 3) = 0

⇒ (x + 3)(2x – 1) = 0

Either x + 3 = 0, then x = - 3

Or 2x – 1= 0, then 2x = 1

⇒ x = 1/2

∴ x = - 3, 1/2

4. (i) 3x2 – 5x – 12 = 0

(ii) 21x2 – 8x – 4 = 0

(i) 3x2 – 5x – 12 = 0

⇒ 3x2 – 9x + 4x – 12 = 0

⇒ 3x(x – 3) + 4(x – 3 ) = 0

⇒ (x – 3)(3x + 4) = 0

Either x – 3 = 0, then x = 3

Or 3x + 4 = 0, then 3x = - 4

⇒ x = -4/3

Hence x = 3, -4/3

(ii) 21x2 – 8x – 4 = 0

21x2 – 8x – 4 = 0

⇒ 21x2 – 14x + 6x – 4 = 0

⇒ 7x(3x – 2) + 2(3x – 2) = 0

⇒ (3x – 2)(7x + 2) = 0

Either 3x – 2 = 0 , then 3x = 2

⇒ x = 2/3

Or 7x + 2 = 0, then 7x = - 2

⇒ x = -2/7

Hence x = 2/3, -2/7

5. (i) 3x2 = x + 4

(ii) x(6x – 1) = 35

(i) 3x2 = x + 4

⇒ 3x2 – x – 4 = 0

⇒ 3x2 – 4x + 3x – 4 = 0

⇒ x(3x – 4) + 1(3x – 4) = 0

⇒ (3x – 4)(x + 1) = 0

Either 3x – 4 = 0, then 3x = 4

⇒ x = 4/3

Or x +1 = 0, then x = - 1

Hence, x= 4/3, -1

(ii) x(6x – 1) = 35

x(6x – 1) = 35

⇒ 6x2 – x – 35 = 0

⇒ 6x2 – 15x + 14x – 35 = 0

⇒ 3x(2x – 5) + 7(2x – 5) = 0

⇒ (2x – 5)(3x + 7) = 0

Either 2x – 5 = 0, then 2x = 5

⇒ x = 5/2

Or 3x + 7 = 0, then 3x = - 7

⇒ x = -7/3

Hence x = 5/2, - 7/3

6. (i) 6p2 – 11p – 10 = 0

(ii) 2/3x2 – 1/3x = 1

(i) 6p2 + 11p – 10 = 0

⇒ 6p2 + 15p – 4p – 10 = 0

⇒ 3p (2p + 5) – 2(2p + 5) = 0

(2p + 5)(3p – 2) = 0

Either 2p + 5 = 0, then 2p = - 5

⇒ p = -5/2

Or 3p – 2 = 0, then 3p = 2

⇒ p = 2/3

Hence p = -5/2, 2/3

(ii) 2/3.x2 – 1/3.x = 1

2/3.x2 – 1/3.x = 1

⇒ 2x2 – x = 3

⇒ 2x2 – x – 3 = 0

⇒ 2x2 – 3x + 2x – 3 = 0

⇒ x(2x – 3) + 1(2x – 3) = 0

⇒ (2x – 3)(x + 1) = 0

Either 2x – 3 = 0, then 2x = 3

⇒ x = 3/2

Or x + 1 = 0, then x = - 1

Hence x = 3/2, - 1

7. (i) (x – 4)2 + 52 = 132

(ii) 3(x – 2)2 = 147

(i) (x – 4)2 + 52 = 132

x2 – 8x + 16 + 25 = 169

x2 – 8x + 41 – 169 = 0

⇒ x2 – 8x – 128 = 0

⇒ x2 – 16x + 8x – 128 = 0

⇒ x(x – 16) + 8(x – 16) = 0

⇒ (x – 16)(x + 8) = 0

Either x – 16 = 0, then x = 16

Or x + 8 = 0, then x = - 8

Hence x = 16, - 8

(ii) 3(x – 2)2 = 147

3(x – 2)2 = 147

3(x2 – 4x + 4) = 147

⇒ 3x2 - 12x + 12 – 147 = 0

⇒ 3x2 – 12x – 135 =0

⇒ x2 – 4x – 45 = 0 (dividing by 3)

⇒ x2 – 9x + 5x – 45 = 0

⇒ x(x – 9) + 5(x – 9) = 0

⇒ (x – 9)(x + 5) = 0

Either x – 9 = 0, then x = 9

Or x + 5 = 0, then x = - 5

Hence x = 9, - 5

8. (i) 1/7(3x – 5)2 = 28

(ii) 3(y2 – 6) = y(y + 7) – 3

(i) 1/7(3x – 5)2 = 28

(3x – 5)2 = 28 × 7

⇒ 9x2 – 30x + 25 = 196

⇒ 9x2 – 30x + 25 – 196 = 0

⇒ 9x2 – 30x – 171 = 0

⇒ 3x2 – 10x – 57 = 0 (Dividing by 3)

⇒ 3x2 - 19x + 9x – 57 = 0

⇒ x(3x – 19) + 3(3x – 19) = 0

⇒ (3x – 19)(x + 3) = 0

Either 3x – 19 = 0, then 3x = 19

⇒ x = 19/3 or x + 3 = 0, then x = - 3

Hence x = 19/2, - 3

(ii) 3(y2 – 6) = y(y + 7) – 3

3(y2 – 6) = y(y + 7) – 3

⇒ 3(y2 – 6) = y2 + 7y – 3

⇒ 3y2 – 18 = y2 + 7y – 3

⇒ 3y2 – y2 – 7y – 18 + 3 = 0

⇒ 2y2 – 7y – 15 = 0

⇒ 2y2 – 10y + 3y – 15 = 0

2y(y – 5) + 3(y – 5) = 0

⇒ (y – 5)(2y + 3) = 0

Either y – 5 = 0, then y = 5

Or 2y + 3 = 0, then 2y = - 3

⇒ y = -3/2

Hence y = -3/2, 5

9. x2 – 4x – 12 = 0, when x N

x2 – 4x - 12 = 0

⇒ x2 – 6x + 2x – 12 = 0

⇒ x(x – 6) + 2(x – 6) = 0

⇒ (x – 6)(x + 2) = 0

Either x – 6 = 0, then x = 6

Or x + 2 = 0, then x = - 2

But - 2 is not a natural number

∴ x = 6

10. 2x2 – 8x – 24 = 0 when x I

2x2 – 8x – 24 = 0

⇒ x2 – 4x – 12 = 0 (Dividing by 2)

⇒ x2 – 6x + 2x – 12 = 0

⇒ x(x – 6) + 2(x – 6) = 0

⇒ (x – 6)(x + 2) = 0

Either x – 6 = 0, then x = 6

Or x+ 2 = 0, then x = - 2

Hence x= 6, - 2

11. 5x2 – 8x – 4 = 0 when x I

2x2 – 8x – 24 = 0

⇒ x2 – 4x – 12 = 0 (Dividing by 2)

⇒ x2 – 6x + 2x – 12 = 0

⇒ x(x – 6) + 2(x – 6) = 0

⇒ x(x – 6)(x + 2) = 0

Either x – 6 = 0, then, x = 6

Or x + 2 = 0, then x = - 2

Hence x = 6, - 2

12. 5x2 – 8x – 4 = 0 when x Q

5x2 – 8x – 4 = 0

∵ 5×(-4) = - 20

⇒ - 20 = - 10 + 2

⇒ - 8 = - 10 + 2

⇒ - 8 = - 10 + 2

⇒ 5x2 – 10x + 2x – 4 = 0

⇒ 5x (x – 2) + 2(x – 2) = 0

⇒ (x – 2)(5x + 2) = 0 (zero Product Rule)

Either x – 2 = 0, then x = 2

or 5x + 2 = 0, then 5x = - 2

⇒ x = - (2/5)

∴ x = 2, - (2/5)

13. 2x2 – 9x + 10 = 0, when

(i) x N

(ii) x Q

2x2 – 9x + 10 = 0

⇒ 2x3 – 4x – 5x + 10 = 0

⇒ 2x(x – 2) – 5(x – 2) = 0

⇒ (x – 2)(2x – 5) = 0

Either x – 2 = 0, then x = 2,

Or 2x – 5 = 0, then 2x = 5

⇒ x = 5/2

(i) when x ∈ N, then x = 2

(ii) when x ∈ Q, then x = 2, 5/2

13. (i) a2x2 + 2ax + 1 = 0, a ≠ 0

(ii) x2 – (p + q)x + pq = 0

(i) a2x2 + 2ax + 1 = 0

⇒ a2x2 + ax + ax + 1 = 0

⇒ ax(ax + 1) + 1(ax + 1) = 0

⇒ (ax + 1)(ax + 1) = 0

⇒ (ax + 1)2 = 0

∴ ax + 1 = 0

⇒ ax = - 1,

x = -1/a

Hence x = -(1/a), - (1/a)

(ii) x2 – (p + q)x + pq = 0

x2 – (p + q)x + pq = 0

x2 – px – qx + pq = 0

x(x – p) – q(x – p) = 0

⇒ (x – p)(x – q) = 0

Either x – p = 0, then x = p,

Or x – q = 0, then x = q

Hence x = p, q

14. a2x2 + (a2 + b2)x + b2 = 0, a ≠ 0

a2x2 + (a2 + b2)x + b2 = 0

⇒ a2x(x + 1) + b2(x + 1) = 0

⇒ (x + 1)(a2x + b2) = 0

⇒ (x + 1) = 0, then x = - 1

Or, a2x + b2 = 0, then a2x = - b2

⇒ x = -b2/a2

Hence x = -1, -b2/a2

15. (i) √3x2 + 10x + 7√3 = 0

(ii) 4√3x2 + 5x - 2√3 = 0

(i) √3x2 + 10x + 7√3 = 0

[∵ √3 × 7√3 = 7×3 = 21]

⇒ √3x(x + √3) + 7(x + √3) = 0

⇒ (x + √3)( ⇒3x + 7) = 0

Either x + √3 = 0, then x = -√3

or 3x + 7 = 0, then x = - √3

⇒ 3x = - 7

⇒ x = -7/√3

⇒ x = (-7 ×√3)/(√3 ×√3)

= (-7√3)/3

Hence, x = - √3, - 7√3/3

(ii) 4√3x2 + 5x - 2√3 = 0

4√3x2 + 5x - 2√3 = 0

{4√3 × (-2√3) = 8 × (-3) = - 24}

4√3x2 + 8x – 3x - 2√3 = 0

⇒ 4x(√3x + 2) - √3(√3x + 2) = 0

⇒ (√3x + 2)(4x - √3) = 0

Either √3x + 2 = 0, then √3x = - 2

⇒ x = - 2/√3

⇒ x = (-2 ×√3)/(√3 ×√3)

= (-2√3)/3

Or, 4x - √3 = 0, then 4x = √3

⇒ x = √3/4

Hence x = (- 2/√3)/3, √3/4

16. (i) x2 – (1 + √2)x + √2 = 0

(ii) x + 1/x = 2.1/20

(i) x2 + (1 + √2)x + √2 = 0

⇒ x2 – x - √2x + √2 = 0

⇒ x(x – 1) - √2(x – 1) = 0

⇒ (x – 1)(x - √2) = 0

Either x – 1 = 0, then x = 1

Or x - √2 = 0, then x = √2

Hence x = 1, √2

(ii) x + 1/x = 2.1/20

(x2 + 1)/x = 41/20

⇒ 20x2 + 20 = 41x

⇒ 20x2 – 16x – 25x + 20 = 0

⇒ 4x(5x – 4) – 5(5x – 4) = 0

⇒ (5x – 4)(4x – 5) = 0

Either 5x – 4 = 0, then 5x = 4

⇒ x = 4/5

or 4x – 5 = 0, then 4x = 5

⇒ x = 5/4

Hence, x = 4/5, 5/4

17. (i) 2/x2 – 5/x + 2 = 0, x ≠ 0

(ii) x2/15 – x/3 – 10 = 0

(i) 2/x2 – 5/x + 2 = 0, x ≠ 0

⇒ 2 – 5x + 2x2 = 0

⇒ 2x2 – 5x + 2 = 0

{∵ 2 × 2 = 4, 4 = - 4 × (-1) , - 5 = - 4 – 1}

⇒ 2x2 – 4x – x + 2 = 0

⇒ 2x(x – 2) – 1(x – 2) = 0

⇒ (x – 2)(2x – 1) = 0

Either x – 2 = 0, then x = 2

Or 2x – 1 = 0, then 2x = 1

⇒ x = 1/2

∴ x = 2, 1/2

(ii) x2/15 – x/3 – 10 = 0

⇒ x2 – 5x – 150 = 0 {∵ - 150 = - 15 × 10, - 5 = - 15 + 10}

⇒ x2 – 15x + 10x – 150 = 0

⇒ x(x – 15) + 10(x – 15) = 0

⇒ (x – 15)(x + 10) = 0

Either x – 15 = 0, then x = 15

Or x + 10 = 0, then x = - 10

∴ x = 15, - 10

18. (i) 3x – 8/x = 2

(ii) (x + 2)/(x + 3) = (2x – 3)/(3x – 7)

(i) 3x – 8/x = 2

(3x2 – 8)x = 2

⇒ 3x2 - 8 = 2x

⇒ 3x2 – 2x – 8 = 0

⇒ 3x2 – 6x + 4x – 8 = 0

⇒ 3x(x – 2) + 4(x – 2) = 0

⇒ (x – 2)(3x + 4) = 0

Either x – 2 = 0, then x = 2

Or 3x + 4 = 0, then 3x = - 4

⇒ x = -4/3

Hence, x = 2, -4/3

(ii) (x + 2)/(x + 3) = (2x – 3)/(3x – 7)

(x + 2)(3x – 7) = (2x – 3)(x + 3)

⇒ 3x2 – 7x + 6x – 14 = 2x2 + 6x – 3x – 9

⇒ 3x2 – x – 14 = 2x2 + 3x – 9

⇒ 3x2 – x – 14 – 2x2 – 3x + 9 = 0

⇒ x2 – 4x – 5 = 0

⇒ x2 – 5x + x – 5 = 0

x(x – 5) + 1(x – 5) = 0

⇒ (x – 5)(x + 1) = 0

Either x – 5 = 0, then x = 5

Or x + 1 = 0, then x = - 1

Hence x = 5, - 1

19. (i) 8/(x + 3) – 3/(2 – x) = 2

(ii) x/(x – 1) + (x – 1)/x = 2.1/2

(i) 8/(x + 3) – 3/(2 – x) = 2

(16x – 8x – 3x – 9)/(x + 3)(2 – x) = 2

⇒ (- 11x + 7)/(2x – x2 + 6 – 3x) = 2

⇒ - 11x + 7 = 4x – 2x2 + 12 – 6x

⇒ - 11x + 7 – 4x + 2x2 – 12 + 6x = 0

⇒ 2x2 – 9x – 5 = 0

⇒ 2x2 – 10x + x – 5 = 0

⇒ 2x(x – 5) + 1(x – 5) = 0

⇒ (x – 5)(2x + 1) = 0

Either x – 5 = 0, then x = 5

Or 2x + 1 = 0, then 2x = -1

⇒ x – 1/2

Hence x = 5, - 1/2

(ii) x/(x – 1) + (x – 1)/x = 2.1/2

x/(x – 1) + (x – 1)/x = 2.1/2

x/(x – 1) + (x – 1)/x = 5/2

⇒ (x2 + x2 – 2x + 1)/x(x – 1) = 5/2

⇒ (2x2 – 2x + 1)/(x – x) = 5/2

⇒ 4x2 – 4x + 2 = 5x – 5x

⇒ 4x2 – 4x + 2 – 5x2 + 5x = 0

⇒ - x2 + x + 2 = 0

⇒ x2 – x – 2 = 0

⇒ x2 – 2x + x – 2 = 0

⇒ x(x – 2) + 1(x – 2) = 0

⇒ (x – 2)(x + 1) = 0

Either x – 2 = 0, then x = 2

or x + 1 = 0, then x = - 1

Hence x = 2, - 1

20. (i) x/(x + 1) + (x + 1)/x = 34/15

(ii) (x + 1)/(x – 1) + (x – 2)/(x + 2) = 3

(i) x/(x + 1) + (x + 1)/x = 34/15

(x2 + x2 + 2x + 1)/x(x + 1) = 34/15

⇒ (2x2 + 2x + 1)/(x2 + x) = 34/15

⇒ 30x2 + 30x + 15 = 34x2 + 34x

⇒ 30x2 + 30x + 15 – 34x2 – 34x = 0

⇒ - 4x2 – 4x + 15 = 0

⇒ 4x2 + 4x – 15 = 0

⇒ 4x2 + 4x – 15 = 0

⇒ 4x2 – 10x – 6x – 15 = 0

⇒ 2x(2x + 5) – 3(2x + 5) = 0

⇒ (2x + 5)(2x – 3) = 0

Either 2x + 5 = 0, then 2x = - 5

⇒ x = -5/2

Or 2x – 3 = 0, then 2x = 3

⇒ x = 3/2

Hence x = -5/2, 3/2

(ii) (x + 1)/(x – 1) + (x – 2)/(x + 2) = 3

(x + 1)/(x – 1) + (x – 2)/(x + 2) = 3

⇒ (x + 1)(x + 2) + (x – 2)(x – 1)/(x – 1)(x + 2) = 3

⇒ (x2 + 2x + x + 2 + x2 – x – 2x + 2)/(x2 + 2x – x – 2)

⇒ (x2 + 3x + 2 + x2 – 3x + 2)/(x2 + x – 2) = 3/1

⇒ 2x2 + 4 = 3x2 + 3x – 6

⇒ 2x2 + 4 – 3x2 – 3x + 6 = 0

⇒ - x2 – 3x + 10 = 0

⇒ x2 + 3x + 10 = 0

⇒ x2 + 3x – 10 = 0

⇒ x2 + 5x – 2x – 10 = 0

⇒ x(x + 5) – 2(x + 5) = 0

⇒ (x + 5)(x – 2) = 0

Either x + 5 = 0, then x = - 5

Or x – 2 = 0, then x = 2

Hence x = - 5, 2

21. (i) 1/(x – 3) – 1/(x + 5) = 1/6

(ii) (x – 3)/(x + 3) + (x + 3)/(x – 3) = 2.1/2

(i) (x + 5 – x + 3)/(x – 3)(x + 5) = 1/6

⇒ 8/(x2 + 2x – 15) = 1/6

⇒ x2 + 2x – 15 = 48

⇒ x2 + 2x – 15 – 48 = 0

⇒ x2 + 2x – 63 = 0

⇒ x2 + 9x – 7x – 63 = 0

⇒ x(x + 9) – 7(x + 9) = 0

⇒ (x + 9)(x – 7) = 0

Either x + 9 = 0, then x = - 9

Or x – 7 = 0, then x = 7

Hence x = - 9, 7

(ii) (x – 3)/(x + 3) + (x + 3)/(x – 3) = 2.1/2

(x – 3)/(x + 3) + (x + 3)/(x – 3) = 2.1/2

Put (x – 3)/(x + 3) = a, then (x + 3)/(x – 3) = 1/ a

∴ a + 1/a = 5/2

2a2 + 2 = 5a

⇒ 2a2 – 5a + 2 = 0

⇒ 2a2 – a – 4a + 2 = 0

⇒ a(2a – 1) – 2(2a – 1) = 0

⇒ (2a – 1)(a – 2) = 0

Either 2a – 1 = 0, then a = 1/2

Or a – 2 = 0, then a = 2

(a) when a = 1/2, then

(x – 3)/(x + 3) = 1/2

⇒ 2x – 6 = x + 3

⇒ 2x – x = 3 + 6

⇒ x = 9

(b) When a = 2, then

(x – 3)/(x + 3) = 2/1

⇒ 2x - 6 = x + 3

⇒ 2x – x = - 3 – 6

⇒ x = - 9

∴ x = 9, - 9

22. (i) a/(ax – 1) + b/(bx – 1) = a + b, a + b ≠ 0, ab ≠ 0

(ii) 1/(2a + b + 2x) = 1/2a + 1/b + 1/2x

(i) a/(ax – 1) + b/(bx – 1) = a + b

⇒ {a/(ax – 1) – b} + {b/(bx – 1) – a} = 0

⇒ (a – abx + b)/(ax – 1) + {b – abx + a)/(bx – 1) = 0

⇒ (a + b – abx)[1/(ax – 1) + 1/(bx – 1)] = 0

⇒ (a + b – abx)[(bx – 1 + ax – 1)/(ax – 1)(bx – 1)] = 0

⇒ (a + b – abx)(ax + bx – 2)/(ax – 1)(bx – 1) = 0

⇒ (a + b – abx)(ax + bx – 2) = 0

⇒ Either a + b – abx = 0, then a + b = abx

x = (a + b)/ab

or ax + bx – 2 = 0, then x(a + b) = 2

x = 2/(a + b)

Hence x = (a + b)/ab, 2/(a + b)

(ii) 1/(2a + b + 2x) = 1/2z + 1/b + 1/2x

⇒ 1/(2a + b + 2x) – 1/2x = 1/2a + 1/b

⇒ (2x – (2a + b + 2x)/(2a + b + 2x)2x = (b + 2a)/2ab

⇒ - (2a + b)/(2a + b + 2x)2x = (2a + b)/2ab

⇒ -1/(2a + b + 2x)2x = 1/2ab

⇒ - 2ab = (2a + b + 2x)2x

⇒ 4ax + 2xb + 4x2 = - 2ab

⇒ 4x2 + 2bx + 4ax + 2ab = 0

⇒ 2x(2x + b) + 2a(2x + b) = 0

⇒ (2x + 2a)(2x + b) = 0

⇒ 2x + 2a = 0 or 2x + b = 0

x = - a or x = -b/2

Hence, the roots of the given equations are – a and –b/2

23. 1/(x + 6) + 1/(x – 10) = 3/(x – 4) s

1/(x + 6) + 1/(x – 10) = 3/(x – 4)

⇒ (x – 10 + x + 6)/(x + 6)(x – 10) = 3/(x – 4)

⇒ (2x – 4)/(x + 6)(x – 10) = 3/(x – 4)

⇒ (2x – 4)(x – 4) = 3(x + 6)(x – 10)

⇒ 2x2 – 8x – 4x + 16 = 3(x2 – 4x – 60)

⇒ 2x2 – 8x – 4x + 16 = 3x2 – 12x – 180

⇒ 2x2 – 12x + 16 – 3x2 + 12x + 180 = 0

⇒ - x2 + 196 = 0

⇒ x2 – 196 = 0

⇒ (x)2 – (14)2 = 0

⇒ (x + 14)(x – 14) = 0

Either x + 14 = 0, then x = - 14

or x – 14 = 0, then x = 14

∴ x = 14, - 14

24: Squaring on both sides

3x + 4 = x2

⇒ x2 – 3x – 4 = 0

⇒ x2 – 4x + x – 4 = 0

⇒ x(x – 4) + 1(x – 4) = 0

⇒ (x – 4)(x + 1) = 0

Either x – 4 = 0, then x = 4

or x + 1 = 0, then x = - 1

∴ x = 4, - 1

Check (i)

If x = 4, then R.H.S = x = 4

∴ L.H.S = R.H.S.

Hence x = 4 is its root

If x = - 1, then = √1

= 1

R.H.S = x = - 1

∵ L.H.S. ≠ R.H.S.

∴ x = - 1 is not its root, hence x = 4

(ii) x(x – 7) = 9 × 2

⇒ x2 – 7x = 18

⇒ x2 – 7x – 18 = 0,

⇒ x2 – 9x + 2x – 18 = 0

⇒ x(x – 9) + 2(x – 9) = 0

⇒ (x – 9)(x + 2) = 0

Either x – 9 = 0, then x = 9

or x + 2 = 0, then x = - 2

∴ x = 9, - 2

Check : If x = 9, then = R.H.S

x = 9 is a root

If x = - 2, then ∴ x = - 2 is also its root

Hence x = 9, - 2

25. Use the substitution = 3x + 1 to solve for x : 5(3x + 1)2 + 6(3x + 1) – 8 = 0

y = 3x + 1

Now, 5(3x + 1)2 + 6(3x + 1) – 8 = 0

Substituting the value of 3x + 1, we get

Substituting the value of 3x + 1, we get

5y2 + 6y – 8 = 0

⇒ 5y2 + 19y – 4y – 8 = 0

{∵ 5 × (- 8) = - 40, ∴ - 40 = 10 × (-4), 6 = 10 – 4 }

⇒ 5y (y + 2) – 4(y + 2) = 0

⇒ (y + 2)(5y – 4) = 0

Either y + 2 = 0, then 5y = 4 ⇒ y = 4/5

(i) If y = - 2, then

3x + 1 = 4/5

⇒ 3x = 4/5 – 1 = -1/5

⇒ x = - 1/5 × 1/3 = -/15

Hence x = - 1, -1/15

26. Find the values of x if P + 1 = 0 and x2 + px – 6 = 0

p + 1 = 0, then p = - 1

Substituting the value of p in the given quadratic equation

x2 + (-1)x – 6 = 0

⇒ x2 – x – 6 = 0

⇒ x2 – 3x + 2x – 6 = 0

⇒ x(x – 3) + 2(x – 3) = 0

⇒ (x – 3)(x + 2) = 0

Either x – 3 = 0, then x = 3

Or x + 2 = 0, then x = - 2

Hence x = 3, - 2

27. Find the values of x if p + 7 = 0, q – 12 = 0 and x2 + px + q = 0

p + 7 = 0, then p = - 7

and q – 12 = 0, then q = 12

Substituting the values of p and q in the given quadratic equation,

x2 – 7x + 12 = 0

⇒ x2 – 3x – 4x + 12 = 0

⇒ x(x – 3) – 4(x – 3) = 0

⇒ (x – 3)(x – 4) = 0

Either x – 3 = 0, then x = 3

Or x – 4 = 0, then x = 4

Hence, x = 3, 4

28. If x = p is a solution of the equation x(2x + 5) = 3, then find the value of p.

Given, x = p and x(2x + 5) = 3

Substituting the value of p, we get

P(2p + 5) = 3

⇒ 2p2 + 5p – 3 = 0

⇒ 2p2 + 6p – p – 3 = 0

{∵ 2 × (- 3) = - 6, ∴ - 6 = 6 × (- 1), 5 = 6 – 1}

⇒ 2p(p + 3) – 1(p + 3) = 9

⇒ (p + 3)(2p – 1) = 0

Either p + 3 = 0, then p = - 3

or 2p – 1 = 0, then 2p = 1

⇒ p = 1/2

∴  p = 1/2, - 3

### Exercise 5.3

Solve the following (1 to 8) equations by using formula:

1. (i) 2x2 – 7x + 6 = 0

(ii) 2x2 – 6x + 3 = 0

(i) 2x2 – 7x + 6 = 0

Here a = 2, b = - 7, c = 6

∴ D = b2 – 4ac = (-7)2 – 4 × 2 × 6

= 49 – 48

= (- b ± √D)/2a

= -(-7) ± √1/(2 × 2)

= (7 ± 1)/4

∴ x1 = (7 + 1)/4 = 8/4 = 2 and x2 = (7 – 1)/4 = 6/4 = 3/2

∴ x = 2, 3/2

(ii) 2x2 – 6x + 3 = 0

Here, a = 2, b = - 6, c = 3

Then D = b2 – 4ac = (-6)2 – 4 × 2 × 3

= 36 – 24

= 12

Now x = (- b ± √D)/2a

= - (- 6) ±2√3)/(2 × 2)

= (6 ± 2√3)/4

∴ x1 = (6 + 2√3)/4 = 2(3 + √3)/4 = (3 + √3)/2

x2 = (6 - 2√3/4)/4 = 2(3 - √3)/4

= ( 3 - √3)/2

Hence, x = (3 + √3)/2 , (3 - √3)/2

2. (i) x2 + 7x – 7 = 0

(ii) (2x + 3)(3x – 2) + 2 = 0

(i) x2 + 7x – 7 = 0

Here a = 1, b = 7, c = - 7

∴ D = b2 – 4ac

= (7)2 – 4 × 1 (-7) = 49 + 28

= 77

(ii) (2x + 3)(3x – 2) + 2 = 0

6x2 – 4x + 9x – 6 + 2 = 0

6x2 + 5x – 4 = 0

Here, a = 6, b = 5, c = - 4

D = b2 – 4ac

= (5)2 – 4 × 6 × (-4)

= 25 + 96

= 121

∵ x = (- b ± √D)/2a ∴ x1 = (- 5 + 11)/12

= 6/12

= 1/2

x2 = (- 5 – 11)/12

= -16/12

= -4/3

Hence x = 1/2, - 4/3

3. (i) 256x2 – 32x + 1 = 0

(ii) 25x2 + 30x + 7 =0

(i) 256x2 – 32x + 1 = 0

Here a = 256, b = - 32, c = 1

D = b2 – 4ac

= (- 32)2 – 4 × 256 × 1

= 1024 – 1024

= 0

∵ x = (- b ± √D)/2a

= - (-32) ± √10)/(2 × 256)

= 32/512

= 1/16

x1 = 1/16, x2 = 1/16

Hence x = 1/16, 1/16

(ii) 25x2 + 30x + 7 = 0

Here a = 25, b = 30, c = 7

D = b2 – 4ac

= (30)2 - 4 × 25 ×7

= 900 – 700

= 200

∵ x = (- b ± √D)/2a

= (- 30 ± 10√2)50

= (- 3 ± √2)/5

∴ x1 = (- 3 + √2)/5 and x2 = (- 3 - √2)/5

Hence x = (- 3 + √2)/5, (- 3 - √2)/5

4. (i) 2x2 + √5x – 5 = 0

(ii) √3x2 + 10x - 8√3 = 0

(i) 2x2 + √5x – 5 = 0

Here a = 2, b = √5, x = - 5

D = b2 – 4ac

= (√5)2 – 4 × 2 × (-5)

= 5 + 40

= 45 x2 = (- √5 - 3√5)/4

= (-4√5)/4

= -√5

Hence x = √5/2, - √5

(ii) √3x2 + 10x - 8√3 = 0

Here a = √3, b = 10, c = - 8√3

D = b2 – 4ac

= (10)2 – 4 × √3 × (-8√3)

= 100 + 96

x2 = (-10 – 14)/(2√3)

= (-24)/(2√3)

= (- 12 ×√3)/( √3 ×√3)

= (- 12√3)/3

= - 4√3

Hence x = (2√3)/3, - 4√3

5. (i) (x – 2)/(x + 2) + (x + 2)/(x – 2) = 4

(ii) (x + 1)/(x + 3) = (3x + 2)/(2x + 3)

(i) (x – 2)/(x + 2) + (x + 2)/(x – 2) = 4

⇒ {(x – 2)2 + (x + 2)2}/{(x + 2)(x – 2)} = 4

⇒ (x2 – 4x + 4 + x2 + 4x + 4)/(x2 – 4) = 4

⇒ 2x2 + 8 = 4x2 – 16

⇒ 2x2 + 8 – 4x2 + 16 = 0

⇒ - 2x2 + 24 = 0

⇒ x2 – 12 = 0

Here a = 1, b = 0, c = - 12

D = b2 – 4ac

= (0)2 – 4 × 1(-12)

= 0 + 48

= 48

∵ x = (- b ± √D)/2a Hence roots are 2√3, - 2√3

(ii) (x + 1)/(x + 3) = (3x + 2)/(2x + 3)

(x + 1)(2x + 3) = (3x + 2)(x + 3)

⇒ 2x2 + 3x + 2x + 3 = 3x2 + 9x + 2x + 6

⇒ 2x2 + 5x + 3 – 3x2 – 11x – 6 = 0

⇒ - x2 – 6x – 3 = 0

⇒ x2 + 6x + 3 = 0

Here a = 1, b = 6, c = 3

D = b2 – 4ac

= (6)2 – 4 × 1 × 3

= 36 – 12

= 24

∵ x = (- b ± √D)/2a ∴ x1 = - 3 + √6, x2 = - 3 - √6

Hence, x = - 3 + √6, - 3 - √6

6. (i) a(x2 + 1) = (a2 + 1)x, a ≠ 0

(ii) 4x2 – 4ax + (a2 – b2) = 0

(i) a(x2 + 1) = (a2 + 1)x

ax2 – (a2 + 1) x + a = 0

Here a = a, b = - (a2 + 1), c = a

D = b2 – 4ac

= [- (-a2 + 1)2 – 4 × a × a

= a4 + 2a2 + 1 – 4a2

= a4 – 2a2 + 1

= (a2 – 1)2

∵ x = (- b ± √D)/2a

∴ x1 = (a2 + 1 + a2 – 1)/2a

= (2a2)/2a = a

x2 = (a2 + 1 – a2 + 1)/2a

= 2/2a

= 1/a

Hence x = a, 1/a

(ii) 4x2 – 4ax + (a2 – b2) = 0

Here a = 4, b = - 4a, c = a2 – b2

D = b2 – 4ac

= (-4a)2 - 4 × 4(a2 – b2)

= 16a2 – 16(a2 – b2)

= 16a2 – 16a2 + 16b2

D = 16b2

∴ x1 = (a + b)/2, x2 = (a – b)/2

Hence x = (a + b)/2, (a – b)/2

7. (i) x – (1/x) = 3, x ≠ 0

(ii) (1/x) + 1/(x – 2) = 3, x ≠ 0, 2

(i) x – 1/x = 3

x2 – 1 = 3x

⇒ x2 – 3x – 1 = 0

Here a = 1, b = - 3, c = - 1

∴ b2 – 4ac

= (3)2 – 4 × 1 × (-1)

= 9 + 4

= 13 (ii) 1/x + 1/(x – 2) = 3

(x – 2 + x)/{x(x – 2)} = 3

⇒ (2x – 2)/(x2 – 2x) = 3

⇒ 3x2 – 6x = 2x - 2

⇒ 3x2 – 6x – 2x + 2 = 0

⇒ 3x2 – 8x + 2 = 0

Here a = 3, b = - 8, c = 2

b2 – 4ac

= (-8)2 – 4 × 3 × 2

= 64 – 24

= 40 8. 1/(x – 2) + 1/(x – 3) + 1/(x – 4) = 0

1/(x – 2) + 1/(x – 3) + 1/(x – 4) = 0

⇒ 1/(x – 2) + 1/(x – 3) = 1/(x – 4)

⇒ (x – 3 + x – 2)/{(x – 2)(x – 3) = - 1/(x – 4)

⇒ (2x – 5)/(x2 – 5x + 6) = -1/(x – 4)

⇒ (2x – 5)/(x2 – 5x + 6) = - {1/(x – 4)}

(2x – 5)(x – 4) = - 1(x2 – 5x + 6)

⇒ 2x2 – 8x – 5x + 20 = - x2 + 5x – 6 = 0

⇒ 2x2 – 8x – 5x + 20 + x2 – 5x + 6 = 0

⇒ 3x2 – 18x + 26 = 0

Here, a = 3, b = - 18, c = 26 = (18 ± 2√3)/6

= (9 ± √3)/3 (dividing by 2)

∴ x = (9 + √3)/3, (9 - √3)/3

= 3 + (√3/3), 3 – (√3/3)

= (3 + 1/√3), (3 – 1/√3)

9. Solve for x : 2(2x – 1)/(x + 3) – 3(x + 3)/(2x – 1) = 5, x ≠ - 3, 1/2

x : 2(2x – 1)/(x + 3) – 3(x + 3)/(2x – 1) = 5

Let (2x – 1)/(x + 3) = y then (x + 3)/(2x – 1) = 1/y

∴ 2y – 3/y = 5

⇒ 2y2 – 3 = 5y

⇒ 2y2 – 5y – 3 = 0

Here, a = 2, b = - 5, c = - 3

b2 – 4ac

= (-5)2 – 4 × 2 × (-3)

= 25 + 24

= 49 = (5 ± 7)/4

y = (5 + 7)/4

= 12/4

= 3

Or y = (5 – 7)/4

= (- 2)/4

= (- 1)/2

∴ y = 3, -1/2

When y = 3, then (2x – 1)/(x + 3) = 3

⇒ 3x + 9 = 2x – 1

⇒ 3x – 2x = - 1 – 9

⇒ x = - 10

When y = -1/2, then

Or (2x – 1)/(x + 3) = - 1/2

4x – 2 = - x – 3

4x + x = - 3 + 2

⇒ 5x = - 1

x = - 1/5

∴ x = - 10, -1/5

10. Solve the following equation by using quadratic equations for x and give your

(i) x2 – 5x – 10 = 0

(ii) 5x(x + 2) = 3

(i) x2 – 5x – 10 = 0

On comparing with, ax2 + bx + c = 0

a = 1, b = - 5, c = - 10 = (5 ± 8.06)/2

Either x = (5 + 8.06)/2

= (13.06)/2

∴ x = 6.53, x = - 1.53

(ii) 5x(x + 2) = 3

5x(x + 2) = 3

⇒ 5x2 + 10x = 3

⇒ 5x2 + 10x – 3 = 0

Here a = 5, b = 10, c = - 3

D = b2 – 4ac

= (10)2 – 4 × 5 × (-3)

= 100 + 60

= 160 = {- 10 ± 4(3.162)}/10

= (- 10 ± 12.648)/10

∴ x1 = (- 10 + 12.648)/10

= (2.648)/10

= 0.2648

= 0.265

x2 = (- 10 – 12.648)/10

= (- 22.648)/10

= - 2.2648

∴ x = 0.26, - 2.26

11. Solve the following equations by using quadratic formula and give your answer correct to 2 decimal places:

(i) 4x2 – 5x – 3 = 0

(ii) 2x – 1/x = 1

(i) Given equation 4x2 – 5x – 3 = 0

Comparing with ax2 + bx + c = 0, we have

a = 4, b = - 5, c = - 3 = (5 ± 8.544)/8

= (5 + 8.544)/8 or (5 – 8.544)

= (13.544)/8 or (-3.544)/8

= 1.693 or – 0.443

= 1.69 or – 0.44 (correct to 2 decimal places)

(ii) 2x – 1/x = 7

⇒ 2x2 – 7x – 1 = 0 ….(i)

Comparing (i) with ax2 + bx + c, we get,

a = 2, b = - 7, c = - 1 ⇒ x = (7 + 7.55)/4 or x = (7 – 7.55)/4

⇒ x = (14.55)/4 or x = (-0.55)/4

⇒ x = 3.64 or x = - 0.14

12. Solve the following equation : x – 18/x = 6. Give your answer correct to two x significant figures.

x – 18/x = 6

⇒ x2 – 6x – 18 = 0

a = 1, b = - 6, c = - 18 = 3 × 2.73 or 3 × - 0.73

= 8.19 or – 2.19

13. Solve the equation 5x2 – 3x – 4 = 0 and give your answer correct to 3 significant figures:

We have 5x2 – 3x – 4 = 0

Here a = 5, b = - 3, c = - 4 x = (3 + 9.43)/10 or x = (3 – 9.43)/10

⇒ x = (12.43)/10 or x = (- 6.43)/10

⇒ x = 1.24 or x = - 0.643

### Exercise 5.4

1. Find the discriminant of the following equations and hence find the nature of roots:

(i) 3x2 – 5x – 2 = 0

(ii) 2x2 – 3x + 5 = 0

(iii) 7x2 + 8x + 2 = 0

(iv) 3x2 + 2x – 1 = 0

(v) 16x2 – 40x + 25 = 0

(vi) 2x2 + 15x + 30 = 0

(i) 3x2 – 5x – 2 = 0

Here a = 3, b = - 5, c = - 2

∴ D = b2 – 4ac

= (5)2 – 4 × 3 × (-2)

= 25 + 24

= 49

∴ Discriminant = 49

∵ D > 0

∴ Roots are real and distinct

(ii) 2x2 – 3x + 5 = 0

Here a = 2, b = - 3, c = 5

∴ D = b2 – 4ac

= (-3)2 – 4 × 2 × 5

= 9 – 40

= - 31

∴ Discriminant = - 31

∵ D < 0

∴ Roots are not real.

(iii) 7x2 + 8x + 2 = 0

Here a = 7, b = 8, c = 2

∴ D = b2 – 4ac

= (8)2 - 4 × 7 × 2

= 64 – 56

= 8

∴ Discriminant = 8

∵ D > 0

∴ Roots are real and distinct

(iv) 3x2 + 2x – 1 = 0

Here a = 3, b = 2, c = - 1

∴ D = b2 – 4ac

= (2)2 – 4 × 3 × (-1)

= 4 + 12

= 16

∴ Discriminant = 16

∵ D > 0

∴ Roots are real and distinct

(v) 16x2 – 40x + 25 = 0

a = 16, b = - 40, c = 25

∴ D = b2 – 4ac

= (-40)2 – 4 × 16 × 25

= 1600 – 1600

= 0

∴ Discriminant = 0

∵ D = 0

∴ Roots are real and equal.

(vi) 2x2 + 15x + 30 = 0

Here, a = 2, b = 15, c = 30

∴ D = b2 – 4ac

= (15)2 – 4 × 2 × 30

= 225 – 240

= - 15

∴ Discriminant = - 15

∵ D < 0

∴ Roots are not real.

2. Discuss the nature of the roots of the following quadratic equations:

(i) x2 – 4x – 1 = 0

(ii) 3x2 – 2x + 1/3 = 0

(iii) 3x2 - 4√3x + 4 = 0

(iv) x2 – 1/2x + 4 = 0

(v) – 2x2 + x + 1 = 0

(vi) 2√3x2 – 5x + √3 = 0

(i) x2 – 4x – 1 = 0

Here, a = 1, b = - 4, c = - 1

∴ D = b2 – 4ac

= (-4)2 – 4 × 1 × (-1)

= 16 + 4

= 20

∵ D > 0

Roots are real and distinct

(ii) 3x2 – 2x + 1/3 = 0

Here a = 3, b = - 2, c = 1/3

∴ D = b2 – 4ac

= (-2)2 – 4 × 3 × 1/3

= 4 – 4

= 0

∵ D = 0

∴ Roots are real and equal.

(iii) 3x2 - 4√3x + 4 = 0

Here a = 3, b = - 4√3, c = 4

∴ D = b2 – 4ac

= (-4√3)2 – 4 × 3 × 4

= 48 – 48

= 0

∵ D = 0

∴ Roots are real and equal

(iv) x2 – 1/2.x + 4 = 0

Here a = 3, b = - 4√3, c = 4

∴ D = b2 – 4ac

= (- 4√3)2 – 4 × 3 × 4

= 48 – 48

= 0

∵ D = 0

∴ Roots are real and equal

(v) –2x2 + x + 1 = 0

Here, a = - 2, b = 1, c = 1

D = b2 – 4ac

= (1)2 – 4 × (-2) × 1

= 1 + 8

= 9

∵ D > 0

∴ Roots are real and distinct.

(vi) 2√3x2 – 5x + √3 = 0

Here a = 2√3, b = - 5, c = √3

∴ D = b2 – 4ac

= (5)2 – 4 × 2√3 ×√3

= 25 – 24

= 1

∵ D > 0

∴ Roots are real and distinct.

3. Find the nature of the roots of the following quadratic equations:

(i) x2 – 1/2x – 1/2 = 0

(ii) x2 - 2√3x – 1 = 0

If real roots exist, find them.

(i) x2 – 1/2.x – 1/2 = 0

Here a = 1, b = -1/2, c = - 1/2

∴  D = b2 – 4ac

= (-1/2)2 – 4 × 1 × (-1/2)

= 1/4 + 2

= 9/4

∵  D = 9/4 > 0

∴ Roots are real and unequal.

(ii) x2 - 2√3x – 1 = 0

Here  a = 1, b = - 2√3, c = - 1

∴ D = b2 – 4ac

= (-2√3)2 – 4 × 1 × (-1)

= 12 + 4

= 16

∵  D > 0

∴  Roots are real and unequal.

4. Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots:

(i) px2 – 4x + 3 = 0

(ii) x2 + ( - 2)x + p = 0

(i) px2 – 4x + 3 = 0

Here  a = p,  b = - 4,  c = 3

∴ D = b2 – 4ac

(-4)2 – 4 × p × 3

= 16 – 12p

∵  The roots are equal.

∴ D = 0

⇒ b2 – 4ac = 0

⇒ 16 – 12p = 0

⇒ 12p = 16

⇒ p = 16/12

= 4/3

∴ p = 4/3

(ii) x2 + (p – 3)x + p = 0

Here a = 1, b = (p – 3), c = p

∵  Equation has real and equal roots

∴ b2 – 4ac = 0

⇒ (p – 3)2 – 4(1) (p) = 0

⇒ (p – 3)2 – 4p = 0

⇒ p2 + 9 – 6p – 4p = 0

⇒ p2 – 10p + 9 = 0

⇒  p2 – 9p – p + 9 = 0

⇒ p(p – 9) – 1(p – 9) = 0

⇒ (p – 1)(p – 9) = 0

∴  p = 1, 9

5. Find the value (s) of k for which each of the following quadratic equation has equal roots:

(i) kx2 – 4x – 5 = 0

(ii) (k – 4)x2 + 2(k – 4)x + 4 = 0

(i) kx2 – 4x – 5 = 0

Here a = k, b = -4, c = 5

∴ D = b2 – 4ac

= (-4)2 – 4 × k × (-5)

= 16 + 20k

∵  Roots are equal.

∴ D = 0

⇒ b2 – 4ac = 0

∴ D = 0

∴ 16 + 20k = 0

⇒ 20k = - 16

⇒ k = - 16/20

= -4/5

Hence k = -4/5

(ii) (k – 4)x2 + 2{k – 4)x + 4 = 0

Here a = k – 4, b = 2(k – 4), c = 4

D = b2 – 4ac

= [2(k – 4)2 – 4 × (k – 4) × 4]

= 4(k2 + 16 – 8k) – 16(k – 4)

= 4(k2 – 8k + 16) – 16(k – 4)

= 4[k2 – 8k + 16 – 4k + 16]

= 4 (k2 – 12k + 32)

∵  Roots are equal

∴ D = 0

⇒ 4(k2 – 12k + 32) = 0

⇒ k2 – 12k + 32 = 0

⇒ k2 – 8k – 4k + 32 = 0

⇒ k(k – 8) – 4(k – 8) = 0

⇒ (k – 8)(k – 4) = 0

Either k – 8 = 0, then k = 8

Or k – 4 = 0, then k = 4

Bur k – 4 ≠ 0

k ≠ 4

k = 8

6. Find the value(s) of m for which of the following quadratic equation has real and equal roots:

(i) (3m + 1)x2 + 2(m + 1)x + m = 0

(ii) x2 + 2(m – 1)x + (m + 5) = 0

(i) (3m + 1)x2 + 2(m + 1)x + m = 0

Here, a = 3m + 1, b = 2(m + 1), c = m

∴ D = b2 – 4ac

= [2(m + 1)]2 – 4 × (3m + 1)(m)

= 4(m2 + 2m + 1) – 12m2 – 4m

= 4m2 + 8m + 4 – 12m2 – 4m

= - 8m2 + 4m + 4

∴ Roots are equal.

∴ D = 0

⇒ - 8m2 + 4m + 4 = 0

⇒ 2m2 – m – 1 = 0 (Dividing by 4)

⇒ 2m2 – 2m + m – 1 = 0

⇒ 2m(m – 1) + 1(m – 1) = 0

⇒ (m – 1)(2m + 1) = 0

Either m – 1 = 0, then m = 1

Or 2m + 1 = 0, then 2m = - 1

⇒ m = - (1/2)

(ii) x+ 2 (m - 1) x + (m + 5) = 0

Equating with ax2 + bx + c = 0
a = 1, b = 2 (m - 1), c = (m + 5)
Since equation has real and equal roots.
so, D = 0
⇒ b2 - 4ac = 0
⇒ [2 (m - 1)2 - 4 x 1 x (m + 5) = 0
⇒ 4 (m - 1)2 - 4 (m + 5) = 0
⇒ 4 [m- 2m + 1 - m - 5)] = 0
⇒ m2 - 3m - 4 = 0
⇒ (m + 1) (m - 4) = 0
Either m + 1 = 0
m = -1
or
m - 4 = 0
m = 4
m = -1, 4

7. Find the values of k which each of the following quadratic equation has equal roots:

(i) 9x2 + kx + 1 = 0

(ii) x2 – 2kx + 7k – 12 = 0

Also, find the roots for those values of k in each case.

(i) 9x2 + kx + 1 = 0

Here a = 9, b = k, c = 1

∴ D = b2 – 4ac

= k2 – 4 × 9 × 1

= k2 – 36

∵ D = 0

⇒ k2 – 36 = 0

⇒ (k + 6)(k – 6) = 0

Either k + 6 = 0, then k = - 6

k – 6 = 0, then k = 6

∴ k = 6, -6

(a) If k = 6, then

9x2 + 6x + 1 = 0

⇒ (3x)2 + 2 × 3x × 1 + (1)2 = 0

⇒ (3x + 1)2 = 0

∴ 3x + 1 = 0

⇒ 3x = - 1

x = - (1/3), - (1/3)

(b) If k = - 6, then

9x2 – 6x + 1 = 0

⇒ (3x)2 – 2 × 3x × 1 + (1)2 = 0

⇒ (3x – 1)2 = 0

⇒ 3x – 1 = 0

⇒ 3x = 1

⇒ x = 1/3

x = 1/3, 1/3

(ii) x2 – 2kx + 7k – 12 = 0

Here a = 1, b = - 2k, c = 7k – 12

∴ D = b2 – 4ac

= (- 2k)2 – 4 × 1 × (7k – 12)

= 4k2 – 4(7k – 12)

= 4k2 – 28k + 48

∵ Roots are equal

∴ D = 0

⇒ 4k2 – 28k + 48 = 0

⇒ k2 – 7k + 12 = 0

⇒ k2 – 3k – 4k + 12 = 0

⇒ k(k – 3) – 4(k – 3) = 0

⇒ (k – 3)(k – 4) = 0

Either k – 3 = 0, then k = 3

Or k – 4 = 0, then k = 4

(a) If k = 3, then

x = (- b √D)/2a

= (4k )/(2 × 1)

= (4 × 3)/2

= 12/2

= 6

x = 6, 6

(b) If k = 4, then

x = (- b √D)/2a

= (- (- 2× 4) )/(2 × 1)

= +8/2

= 4

∴ x = 4, 4

8. Find the value(s) of p for which the quadratic equation (2p + 1)x2 – (7p + 2) + (7p – 3) = 0 has equal roots. Also find these roots.

The quadratic equation given is (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0

Comparing with ax2 + bx + c = 0, we have

a = 2p + 1, b = -(7p + 2), c = (7p – 3)

D = b2 – 4ac

⇒ 0 = [- (7p + 2)]2 – 4(2p + 1)(7p – 3)

⇒ 0 = 49p2 + 4 + 28p – 4(14p2 – 6p + 7p – 3)

⇒ 0 = 49p2 + 4 + 28p – 56p2 – 4p + 12

⇒ 0 = - 7p2 + 24p + 16

⇒ 0 = - 7p2 + 28p – 4p + 16

⇒ 0 = - 7p(p – 4) – 4(p – 4)

⇒ 0 = (- 7p – 4)(p – 4) = 0

⇒ - 7p – 4 = 0 or p – 4 = 0

Hence, the value of p = - 4/7 or p = 4

9. If – 5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k.

- 5 is a root of the quadratic equation

2x2 + px – 15 = 0, then

⇒ 2(5)2 – p(-5) – 15 = 0

⇒ 50 – 5p – 15 = 0

⇒ 35 – 5p = 0

⇒ 5p = 35

⇒ p = 35/5 = 7

P(x2 + x) + k = 0 has equal roots

⇒ px2 + px + k = 0

⇒ 7x2 + 7x + k = 0

Here, a = 7, b = 7, c = k

b2 – 4ac

= (7)2 – 4 × 7 × k

= 49 – 28k

∵ Roots are equal

∴ b2 – 4ac = 0

⇒ 49 – 28k = 0

⇒ 28k = 49

⇒ k = 49/28

= 7/4

∴  k = 7/4

10. Find the value(s) of p for which the equation 2x2 + 3x + p = 0 has real roots.

2x2 + 3x + p = 0

Here, a = 2, b = 3, c = p

b2 – 4ac

= (3)2 – 4 × 2 × p

= 9 – 8p

∵ Roots are equal

∴ b2 – 4ac ≥ 0

⇒ 9 – 8p ≥ 0

⇒ 9 ≥ 8p

⇒ 8p ≤ 9

⇒ p ≤ 9/8

11. Find the least positive value of k for which the equation x2 + kx + 4 = 0 has real roots.

x2 + kx + 4 = 0

here, a = 1, b = k, c = 4

b2 – 4ac

= k2 – 4 × 1 × 4

= k2 – 16

∵ Roots are real and positive.

∴ k2 – 16 ≥ 0

⇒ k2 ≥ 16

⇒ k ≥ 4

⇒ k = 4

12. Find the values of p for which the equation 3x2 – px + 5 = 0 has real roots.

3x2 – px + 5 = 0

Here, a = 3, b = - p, c = 5

∴ b2 – 4ac

= (-p)2 – 4 × 3 × 5

= p2 – 60

∴ Roots are equal

∴ b2 – 4ac ≥ 0

∴ p2 – 60 ≥ 0

⇒ p2 ≥ 60 ### Exercise 5.5

1.(i) Find two consecutive natural numbers such that the sum of their squares is 61.

(ii) Find two consecutive integers such that the sum of their squares is 61.

(i) Let the first natural number = x

Then second natural number = x + 1

According to the condition, (x)2 + (x + 1)2 = 61

⇒ x2 + x2 + 2x + 1 – 61 = 0

⇒ 2x2 +2x – 60 = 0

⇒ x2 + x – 30 = 0

⇒ x2 + 6x – 5x – 30 = 0

⇒ x(x + 6) – 5(x + 6) = 0

⇒ (x + 6)(x – 5) = 0

Either x + 6 = 0, then x = - 6, or x – 5 = 0, then x = 5

∵ The numbers are positive.

∴ x = - 6 is not possible.

Hence the first natural number = 5 + 1 = 6

(ii) Let first integer = x

Then second integer = x + 1

According to the condition, (x)2 + (x + 1)2 = 61

⇒ x2 + x2 + 2x + 1 = 61

⇒ 2x2 + 2x + 1 – 61 = 0

⇒ 2x2 + 2x – 60 = 0

⇒ x2 + x – 30 = 0 (Dividing by 2)

⇒ x2 + 6x – 5x – 30 = 0

⇒ x(x + 6) – 5(x + 6) = 0

⇒ (x + 6)(x – 5) = 0

Either x + 6 = 0, then x = - 6 or x – 5 = 0, then x = 5

(i) If x = - 6, then

First integer = - 6

And second = - 6 + 1 = - 5

(ii) If x = 5, then

First integer = 5

And second = 5 + 1 = 6

∴ Required integers are = (-6, - 5), (5, 6)

2. (i) If the product of two positive consecutive even integers is 288, find the integers.

(ii) If the product of two consecutive even integers is 224, find the integers.

(iii) Find two consecutive even natural numbers such that the sum of their squares is 340.

(iv) Find two consecutive odd integers such that the sum of their squares is 394.

(i) Let first positive even integer = 2x

Then second even integer = 2x + 2

According to the condition,

2x × (2x + 2) = 288

⇒ 4x2 + 4x – 288 = 0

⇒ x2 + x – 72 = 0 (Dividing by 4)

⇒ x2 + 9x – 8x – 72 = 0

⇒ x(x + 9) – 8(x + 9) = 0

⇒ (x + 9)(x – 8) = 0

Either x + 9 = 0, then x = - 9

But it is not possible as it is negative

Or x – 8 = 0, then x = 8

∴ First even integer = 2x = 2 × 8 = 16

And second even integer = 16 + 2 = 18

(ii) Let first even integer = 2x

Then second even integer = 2x + 2

According to the condition,

2x × (2x + 2) = 224

⇒ 4x2 + 4x – 224 = 0

⇒ x2 + x – 56 = 0

⇒ x2 + 8x – 7x – 56 = 0

⇒ x(x + 8) – 7(x + 8) = 0

⇒ (x + 8)(x – 7) = 0

Either x + 8 = 0, then x = - 8

∴ First even integer = 2 × (-8) = - 16

And second even integer = - 16 + 2 = - 14

Or x – 7 = 0, then x = 7

∴ First even integer = 2x = 2 × 7 = 14

And second even integer = 14 + 2 = 16

(iii) Let first even natural number = 2x

Then second number = 2x + 2

According to the condition,

(2x)2 + (2x + 2)2 = 340

4x2 + 4x2 + 8x + 4 = 340

⇒ 8x2 + 8x + 4 – 340 = 0

⇒ 8x2 + 8x – 336 = 0

⇒ x2 + x – 42 = 0 (Dividing by 8)

⇒ x2 + 7x – 6x – 42 = 0

⇒ x(x + 7) – 6(x + 7) = 0

⇒ (x + 7)(x – 6) = 0

Either x + 7 = 0, then x = - 7

But it is not a even natural number

or x – 6 = 0, then x = 6

∴ First even natural number = 2x = 2 × 6 = 12

And second = 12 + 2 = 14

∴ Numbers are 12, 14

(iv) Let first odd integer = 2x + 1

Then second odd integer = 2x + 3

According to the condition,

(2x + 1)2 + (2x + 3)2 = 394

⇒ 4x2 + 4x + 1 + 4x2 + 12x + 9 = 394

⇒ 8x2 + 16x – 394 + 10 = 0

⇒ 8x2 + 16x – 384 = 0

⇒ x2 + 2x – 48 = 0 (Dividing by 8)

⇒ x2 + 8x – 6x – 48 = 0

⇒ x(x + 8) – 6(x + 8) = 0

⇒ (x + 8)(x – 6) = 0

Either x + 8 = 0, then x = - 8

Or x – 6 = 0, then x = 6

(a) If x = - 8, then first odd integer = 2x + 1

= 2 × (-8) + 1

= - 16 + 1

= - 15

and second integer = - 15 + 2

= - 13

(b) If x = 6, then first odd integer = 2x + 1

= 2 × 6 + 1

= 13

and second integer = 13 + 2 = 15

∴  Required integers are – 15, - 13, or 13, 15

3. The sum of two numbers is 9 and the sum of their squares is 41. Taking one number as x, form ail equation in x and solve it to find the numbers.

Sum of two numbers = 9

Let first number = x

Then second number = 9 – x

Now according to the condition,

(x)2 + (9 – x)2 = 41

⇒ x2 + 81 – 18x + x2 – 41 = 0

⇒ 2x2 – 18x + 40 = 0

⇒ x2 – 9x + 20 = 0 (Dividing by 2)

⇒ x2 – 4x – 5x + 20 = 0

⇒ x(x – 4) – 5(x – 4) = 0

⇒ (x – 4)(x – 5) = 0

Either x – 4 = 0, then x = 4, or x – 5 = 0, then x = 5

(i) If x = 4, then first number = 4

And second number = 9 – 4 = 5

(ii) If x = 5, then first number = 5

And second number = 9 – 5 = 4

Hence numbers are 4 and 5

4. Five times a certain whole numbers is equal to three less than twice the square of the number. Find the number.

Let number = x

Now according to the condition,

5x = 2 x2 – 3

⇒ 2x2 – 5x – 3 = 0

⇒ 2x2 – 6x + x – 3 = 0

⇒ 2x(x – 3) + 1(x – 3) = 0

⇒ (x – 3)(2x + 1) = 0

Either x – 3 = 0, then x = 3

Or 2x + 1 = 0, then 2x = - 1

⇒ x = -(1/2)

But it is not possible as the number is whole number.

5. Sum of two natural numbers is 8 and the difference of their reciprocal is 2/15. Find the numbers.

Let x and y be two numbers

Given that, x + y = 8 ...(i)

And 1/x – 1/y = 2/15

From equation (i), we have, y = 8 – x

Substituting the value of y in equation (ii),

We have,

1/x – 1/(8 – x) = 2/15

⇒ (8 – x – x)/x(8 – x) = 2/15

⇒ (8 – 2x)/{x(8 – x)} = 2/15

⇒ (4 – x)/{x(8 – x)} = 1/15

⇒ 15(4 – x) = x(8 – x)

⇒ 60 – 15x = 8x – x2

⇒ x2 – 15x – 8x + 60 = 0

⇒ x2 – 23x + 60 = 0

⇒ x2 – 20x – 3x + 60 = 0

⇒ x(x – 20) – 3(x – 20) = 0

⇒ (x – 3)(x – 20) = 0

Either (x – 3) = 0 or (x – 20) = 0

⇒ x = 3 or x = 20

Since sum of two natural numbers is 8 – x

i.e., 8 – 20 cannot be equal to 20

Thus x = 3

From equation (i), y = 8 – x = 8 – 3 = 5

Thus the values of x and y are 3 and 5 respectively.

6. The difference between the squares of two numbers is 45. The square of the smaller number is 4times the larger number. Determine the numbers.

Let the larger number = x

Then smaller number = y

Now according to the condition,

x2 – y2 = 45 .…(i)

and y2 = 4x …(ii)

Substituting the value of y2 from (ii) in (i)

x2 – 4x = 45

⇒ x2 – 4x – 45 = 0

⇒ x2 – 9x + 5x – 45 = 0

⇒ x(x – 9) + 5(x – 9) = 0

⇒ (x – 9)(x + 5) = 0

Either x – 9 = 0, their x = 9, or x + 5 = 0, then x = - 5

(i) When x = 9, the larger number = 9

and smaller number ∴ y = 6

(ii) When x = - 5, then larger number = - 5

Hence numbers are 6, 9

7. There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. What are the integers?

Let the first integer = x

Then second integer = x + 1

And third integer = x + 2

Now according to the condition,

∴ x2 + (x + 1)(x + 2) = 154

⇒ x2 + x2 + 3x + 2 – 154 = 0

⇒ 2x2 + 3x – 152 = 0

⇒ 2x2 + 19x – 16x – 152 = 0

⇒ x(2x + 19) – 8(2x + 19) = 0

⇒ (2x + 19)(x – 8) = 0

Either 2x + 19 = 0, then 2x = - 19

⇒ x = - (19/2)

But it is not possible as it is not an positive integer.

or x – 8 = 0, then x = 8

∴ Numbers are 8, (8 + 1) = 9 and (8 + 2) = 10

8. (i) Find three successive even natural numbers, the sum of whose squares is 308,

(ii) Find three consecutive odd integers, the sum of whose squares is 83.

(i) Let first even number = 2x

Second even number = 2x + 2

Third even number = 2x + 4

Now according to the condition,

(2x)2 + (2x + 2)2 + (2x + 4)2 = 308

⇒ 4x2 + 4x2 + 8x + 4 + 4x2 + 16x + 16 = 308

⇒ 12x2 + 24x + 20 – 308 = 0

⇒ 12x2 + 24x – 288 = 0

⇒ x2 + 2x – 24 = 0 (Dividing by 12)

⇒ x2 + 6x – 4x – 24 = 0

⇒ x(x + 6) – 4(x + 6) = 0

⇒ (x + 6)(x – 4) = 0

Either x + 6 = 0, then x = - 6

But it is not a natural number, hence not possible.

or x – 4 = 0, then x = 4

∴ First even natural number = 2x = 2 × 4 = 8

Second number = 8 + 2 = 10

And the third number = 10 + 2

= 12

(ii) Let the three numbers be x, x + 2, x + 4

According to statement,

(x)2 + (x + 2)2 + (x + 4)2 = 83

⇒ x2 + x2 + 4x + 4 + x2 + 8x + 16 = 83

⇒ 3x2 + 12x + 20 = 83

⇒ 3x2 + 12x + 20 – 83 = 0

⇒ 3x2 + 12x – 63 = 0

⇒ x2 + 4x – 21 = 0

⇒ x2 + 7x – 3x – 21 = 0

⇒ x(x + 7) – 3(x + 7) = 0

⇒ (x – 3)(x + 7) = 0

Either x – 3 = 0 then x = 3 or x + 7 = 0, then x = - 7

∴ Number will be 3, 3 + 2, 3 + 4 = 3, 5, 7

Or Numbers will be – 7, - 7 + 2, - 7 + 4

= - 7, - 5, - 3

9. In a certain positive fraction, the denominator is greater than numerator by 3. If 1 is subtracted from both the numerator and denominator, the fraction is decreased by 1/14. Find the fraction.

Let the numerator of a fraction = x

Then denominator = x + 3

Then fraction = x/(x + 3)

Now according to the condition,

New fraction = (x – 1)/(x + 3 – 1) = x/(x + 3) – 1/14

⇒ (x – 1)/(x + 2) = (14x – x – 3)/{14(x + 3)}

⇒ (x - 1)/(x + 2) = (13x – 3)/(14x + 42)

⇒ (x – 1)(14x + 42) = (13x – 3)(x + 2)

⇒ 14x2 + 42x – 14x – 42 = 13x2 + 26x – 3x – 6

⇒ 14x2 + 28x – 42 – 13x2 – 23x + 6 = 0

⇒ x2 + 5x – 36 = 0

⇒ x2 + 9x – 4x – 36 = 0

x(x + 9) – 4(x + 9) = 0

⇒ (x + 9)(x – 4) = 0

Either x + 9 = 0, then x = - 9, but it is not possible as the fraction is positive.

or x – 4 = 0, then x = 4

∴ Fraction = x/(x + 3) = 4/(4 + 3) = 4/7

10.The sum of the numerator and denominator of a certain positive fraction is 8. If 2 is added to both the numerator and denominator, the fraction is increased by 4/35. Find the fraction.

Let the denominator of a positive fraction = x

Then numerator = 8 – x

∴ Fraction = (8 – x)/x

According to the condition,

(8 – x + 2)/(x + 2) = (8 – x)/x + 4/35

⇒ (10 – x)/(x + 2) = (8 – x)/x + 4/35

⇒ (10 – x)/(x + 2) - (8 – x)/x = 4/35

⇒ (10x – x2 – 8x + x2 – 16 + 2x)/{x(x + 2)} = 4/35

⇒ (4x – 16)/(x2 + 2x) = 4/35

⇒ 4x2 + 8x = 140x – 560

⇒ 4x2 + 8x – 140x + 560 = 0

⇒ 4x2 – 132x + 560 = 0

⇒ x2 – 33x + 140 = 0

⇒ x2 – 28x – 5x + 140 = 0

⇒ x(x – 28) – 5(x – 28) = 0

⇒ (x – 28)(x – 5) = 0

Either x – 28 = 0, then x = 28, but it is not possible as sum of numerator and denominator is 8.

Or x – 5 = 0, then x = 5

∴ Fraction = (8 – x)/x = (8 – 5)/5 = 3/5

11. A two digit number contains the bigger at ten’s place. The product of the digits is 27 and the difference between two digits is 6. Find the number.

Let unit’s digit = x

Hen tens digit = x + 6

Number = x + 10(x + 6)

= x + 10x + 60

= 11x + 60

According to the condition,

x(x + 6) = 27

⇒ x2 + 6x – 27 = 0

⇒ x2 + 9x – 3x – 27 = 0

⇒ x(x + 9) – 3(x + 9) = 0

⇒(x + 9)(x – 3) = 0

Either x + 9 = 0, then x = - 9, but it is not possible as it is negative.

Or x - 3 = 0, then x = 3

∴ Number = 11x + 60

= 11 × 3 + 60

= 33 + 60

12. A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.

Let 2 - digit number = xy = 10x + y

Reversed digits = yx = 10y + x

According to question,

xy = 6

y = 6/x ….(i)

and 10x + y + 9 = 10y + x

⇒ 10x + 6/x + 9 = 10 × 6/x + x (From (i) y = 6/x)

⇒ 10x2 + 6 + 9x = 60 + x2

⇒ 10x2 – x2 + 9x + 6 – 60 = 0

⇒ 9x2 + 9x – 54 = 0

⇒ x2 = x – 6 = 0

⇒ x2 + 3x – 2x – 6 = 0

⇒ x(x + 3) – 2(x + 3) = 0

⇒ (x – 2)(x + 3) = 0

⇒ x = 2 or – 3 (rejecting – 3)

Putting the value of x in (i)

y = 6/2 = 3

∴ 2-digit = 10x + y

= 10 + y

= 10 × 2 + 3

= 23

13. A rectangle of area 105 cm2 has its length equal to x cm. Write down its breadth in terms of x. Given that the perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.

Perimeter of rectangle = 44 cm

Length + breadth = 44/2 = 22 cm

Let length = 22 – x

According to the condition,

x (22 – x) = 105

⇒ x2 – 22x + 105 = 0

⇒ x2 – 15x – 7x + 105 = 0

⇒ x(x – 15) – 7(x – 15) = 0

⇒ (x – 15)(x – 7) = 0

Either x – 15 = 0, then x = 15

Or x – 7 = 0, then x = 7

∴ Length = 15 cm

And breadth = 22 – 15

= 7 cm

14. A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.

Length of garden = 16 m

And width = 10 m

Let the width of walk = x m

Outer length = 16 + 2x

And outer width = 10 + 2x

Now according to the condition,

(16 + 2x)(10 + 2x) – 16 × 10 = 120

⇒ 160 + 32x + 20x + 4x2 – 160 = 120

⇒ 4x2 + 52x – 120 = 0

⇒ x2 + 13x – 30 = 0 (Dividing by 4)

⇒ x2 + 15x – 2x – 30 = 0

⇒ x(x + 15) – 2(x + 15) = 0

⇒ (x + 15)(x – 2) = 0

Either x + 15 = 0, then x = - 15

But it is not possible.

or x – 2 = 0, then x = 2

15. (i) Harish made a rectangular garden, with its length 5 metres more than its width. The next year, he increased the length by 3 metres and decreased the width by 2 metres. If the area of the second garden was 119 sq.m, was the second garden larger or smaller?

(ii) The length of a rectangle exceeds its breadth by 5 m. If the breadth were doubled and the length reduced by 9 m, the area of the rectangle would have increased by 140 m2. Find its dimensions.

In first case,

Let the length of the garden = x m

Then width = (x – 5) m

Area = l ×b

= x(x – 5) sq. m

In second case,

Length = (x + 3)m

And width = x – 5 – 2

= (x – 7)m

According to the condition,

(x + 3)(x – 7) = 119

⇒ x2 – 7x + 3x – 21 = 119

⇒ x2 – 4x – 21 – 119 = 0

⇒ x2 – 4x – 140 = 0

⇒ x2 – 14x + 10x – 140 = 0

⇒ x(x – 14) + 10(x – 14) = 0

⇒ (x – 14) (x + 10) = 0

Either x – 14 = 0, then x = 14

or x + 10 = 0, then x = - 10, but it is not possible as it is negative.

∴ Length of first garden = 14 m

And width = 14 – 5 = 9 m

Area = l × b

= 14 × 9

= 126 m2

Difference of areas of two rectangles = 126 – 119

= 7 sq. m.

∴ Area of second garden is smaller than the area of the first garden by 7 sq. m.

Let length of the rectangle = x m

Then width = (x – 5) m

Area = x(x – 5) sq. m

In second case,

Length of the second rectangle = x – 9

and width = 2(x – 5)m

Area = (x – 9)2(x – 5) = 2(x – 9)(x – 5) sq. m

According to the condition,

2(x – 9)(x – 5) = x(x – 5) + 140

⇒ 2(x2 – 14x + 45) = x2 – 5x + 140

⇒ 2x2 – 8x + 90 = x2 – 5x + 140

⇒ 2x2 – 28x + 90 – x2  + 5x – 140 = 0

⇒ x2 – 23x – 50 = 0

⇒ x2 – 25x + 2x – 50 = 0

⇒ x(x – 25) + 2(x – 25) = 0

⇒ (x – 25)(x + 2) = 0

Either x – 25 = 0, then x = 25

Or x + 2 = 0, then x = - 2, but it is not possible as it is negative.

∴ Length of the rectangle = 25 m and width

= 25 – 5

= 20 m

16. The perimeter of a rectangular plot is 180 m and its area is 1800 m2. Take the length of the plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the values of length, breadth and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot.

The perimeter of a rectangular field = 180 m

And area = 1800 m2

Let length = x m

But length + breadth = 180/2 = 90 m

∴ Breadth = (90 – x)m

According to the condition,

x(90 – x) = 1800

⇒ 90x – x2 – 1800 = 0

⇒ x2 – 90x + 1800 = 0

⇒ x2 – 60x – 30x + 1800 = 0

⇒ x(x – 60) – 30(x – 60) = 0

⇒ (x – 60)(x – 30) = 0

Either x – 60 = 0, then x = 60

Or x – 30 = 0, then x = 30

∵ Length is greater than its breadth

∴ Length = 60 m

And breadth = 90 – 60

= 30 m

17. The lengths of the parallel sides of a trapezium are (x + 9) cm and (2x – 3) cm and the distance between them is (x + 4) cm. If its area is 540 cm2, find x.

Area of a trapezium = 1/2

(sum of parallel sides) × height

Lengths of parallel sides are (x + 9) and (2x – 3)

And height = (x + 4)

According to the condition,

1/2(x + 9 + 2x – 3) × (x + 4) = 540

⇒ (3x + 6)(x + 4) = 540 × 2

⇒ 3x2 + 12x + 6x + 24 – 1080 = 0

⇒ 3x2 + 18x – 1056 = 0

⇒ x2 + 6x – 352 = 0 (Dividing by 3)

⇒ x2 + 22x – 16x – 352 = 0

⇒ x(x + 22) – 16(x + 22) = 0

⇒ (x + 22)(x – 16) = 0

Either x + 22 = 0, then x = - 22

But it is not possible as it is negative

Or x – 16 = 0, then x = 16

18. If the perimeter of a rectangular plot is 68 m and the length of its diagonal is 26 m, find its area.

Perimeter = 68 m and diagonal = 26 m

Length + breadth = 68/2 = 34 m

Let length = x m

Then breadth = (34 – x)m

According to the condition,

l2 + b2 = h2

(x)2 + (34 – x)2 = (26)2

⇒ x2 + 1156 + x2 – 68x = 676

⇒ 2x2 – 68x + 1156 – 676 = 0

⇒ 2x2 – 68x + 480 = 0

⇒ x2 – 34x + 240 = 0 (Dividing by 2)

⇒ x2 – 24x – 10x + 240 = 0

⇒ x(x – 24) – 10(x – 24) = 0

⇒ (x – 24)(x – 10) = 0

Either x – 24 = 0, then x = 24

Or x – 10 = 0, then x = 10

∵ Length is greater than breadth

∴ Length = 24 m

And breadth = (34 – 24) = 10 m

And area = l × b

= 24 × 10

= 240 m2

19. If the sum of the smaller sides of a right- angled triangle is 17cm and the perimeter is 30 cm, then find the area of the triangle.

The perimeter of the triangle = 30 cm

Let one of the two small sides = x

Then, other side = 17 – x ∴ Length of hypotenuse = perimeter – sum of other two sides

= 30 cm – 17 cm

= 13 cm

According to the problem,

x2 + (17 – x)2 = (13)2 (Pythagoras theorem)

⇒ x2 + 289 + x2 – 34x = 169

⇒ 2x2 – 34x + 289 – 169 = 0

⇒ 2x2 – 34x + 120 = 0

⇒ x2 – 17x + 60 = 0 (Dividing by 2)

⇒ x2 – 12x - 5x + 60 = 0

⇒ x(x – 12) - 5(x – 12) = 0

⇒ (x – 12)(x – 5) = 0

Either x – 12 = 0, then x = 12

or x – 5 = 0, then x = 5

(i) When x = 12, then first side = 12 cm

And second side = 17 – 12 = 5 cm

(ii) When x = 5, then first side = 5

And second side = 17 – 5 = 12

∴ Sides are 5 cm, 12 cm

Now, area of the triangle = (5 × 12)/2

= 60/2

= 30 cm2

20. The hypotenuse of grassy land in the shape of a right triangle is 1 metre more than twice the shortest side, find the sides of the grassy land.

Let the shortest side = x

Hypotenuse = 2x + 1

And third side = x + 7

According to the condition,

(2x + 1)2 = x2 + (x + 7)2

⇒ 4x2 + 4x + 1 = x2 + x2 + 14x + 49

⇒ 4x2 + 4x + 1 – 2x2 – 14x – 49 = 0

⇒ 2x2 – 10x – 48 = 0

⇒ x2 – 5x – 24 = 0 (Dividing by 2)

⇒ x2 – 8x + 3x – 24 = 0

⇒ x(x – 8) + 3(x – 8) = 0

⇒ (x – 8)(x + 3) = 0

Either x – 8 = 0, then x = 8

or x + 3 = 0, then x = - 3, but it is not possible as it is negative.

∴ Shortest side = 8 m

Third side = x + 7

= 8 + 7

= 15 m

And hypotenuse = 2x + 1

= 8 × 2 + 1

= 16 + 1

= 17 m

21. Mohini wishes to fit three rods together in the shape of a right triangle. If the hypotenuse is 2 cm longer than the base and 4 cm longer than the shortest side, find the lengths of the rods.

Let the length of hypotenuse = x cm

then base = (x – 2) cm

and shortest side = x – 4

According to the condition,

(x)2 = (x – 2)2 + (x – 4)2

⇒ x2 = x2 – 4x + 4 + x2 – 8x + 16

⇒ x2 = 2x2 – 12x + 20

⇒ 2x2 – 12x + 20 – x2 = 0

⇒ x2 – 12x + 20 = 0

⇒ x2 – 10x – 2x + 20 = 0

⇒ x(x – 10) – 2(x – 10) = 0

⇒ (x – 10)(x – 2) = 0

Either x – 10 = 0, then x = 10

or x – 2 = 0, then x = 2, but it is not possible as the hypotenuse is the longest side.

∴ Hypotenuse = 10 cm

Base = 10 – 2

= 8 cm

And shortest side = 10 – 4

= 6 cm

22. In a P.T. display, 480 students are arranged in rows and columns. If there are 4 more students in each row than the number of rows, find the number of students in each row.

Total number of students = 480

Let the number of students in each row = x

Then the number of rows = 480/x

According to the condition, x = (480/x) + 4

⇒ x2 = 480 + 4x

⇒ x2 – 4x – 480 = 0

⇒ x2 – 24x + 20x – 480 = 0

⇒ x(x – 24) + 20(x – 24) = 0

⇒ (x – 24)(x + 20) = 0

Either x – 24 = 0 or x + 20 = 0

⇒ x = 24

or x = - 20 which is not possible as it is negative

∴ Number of students in each row = 24

23. In an auditorium, the number of rows are equal to the number of seats in each row. If the number of rows is doubled and number of seats in each row is reduced by 5, then the total numbers of seats is increased of 375. How many rows were there?

Let the number of rows = x

Then no. of seats in each row = x

And total number of seats = x × x = x2

According to the condition,

2x × (x - 5) = x2 + 375

⇒ 2x2 – 10x = x2 + 375

⇒ 2x2 – 10x – x2 – 375 = 0

⇒ x2 – 25x + 15x – 375 = 0

⇒ x(x – 25) + 15(x – 25) = 0

⇒ (x – 25)(x + 15) = 0

Either x – 25 = 0, then x = 25

or x + 15 = 0, then x = - 15, but it is not possible as it is negative.

∴ Number of rows = 25

24. At an annual function of a school, each student gives the gift to every other student. If the number of gifts is 1980, find the number of students.

Let the number of students = x

Then the number of gifts given = x – 1

Total number of gifts = x(x – 1)

According to the condition,

x(x – 1) = 1980

⇒ x2 – x – 1980 = 0

⇒ x2 – 45x + 44x – 1980 = 0

⇒ x(x – 45) + 44(x – 45) = 0

⇒ (x – 45)(x + 44) = 0

Either x – 45 = 0, then x = 45

Or x + 44 = 0, then x = - 44, but it is not possible as it is negative.

Hence number of students = 45

25. A bus covers a distance of 240 km at a uniform speed. Due to heavy rain, its speed gets reduced by 10 km/h and as such it takes two hours longer to cover the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate x.

Distance = 240 km

Let speed of a bus = x km/hr

∴ Time taken = D/S = 240/x hours

Due to heavy rains

Speed of the bus = (x – 10) km/hr

∴ Time taken = 240/(x – 10)

According to the condition,

240/x = 240/(x – 10) – 2

240/(x – 10) – 240/x = 2

(240x – 240x – 2400)/{x(x – 10)} = 2

2400/(x2 – 10x) = 2

⇒ 2400 = 2x2 – 20x

⇒ 2x2 – 20x – 2400 = 0

⇒ x2 – 10x – 1200 = 0

⇒ x2 – 40x + 30x – 1200 = 0

⇒ x(x – 40) + 30(x – 40) = 0

⇒ (x – 40)(x + 30) = 0

Either x – 40 = 0, then x = 40

or x + 30 = 0, then x = - 30 which is not possible, speed being negative.

∴ Speed of bus = 40 km/hr

26. The speed of an express train is x km/hr and the speed of an ordinary train is 12 km/hr less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km, find the speed of the express train.

Let the speed of express train = x lm

Then speed of the ordinary train = (x – 12)km

Time is taken to cover 240 km by the express train = 240/x hours

Time taken to cover 240 km by the ordinary train = 240/(x – 12) hours

According to the condition,

240/(x – 12) – 240/x = 1

⇒ 240[1/(x – 12) – 1/x] = 1

⇒ 240[(x – x + 12)/{x(x – 12)}] = 1

⇒ 240[12/(x2 – 12x)] = 1

⇒ 2880 = x2 – 12x

⇒ x2 – 12x – 2880 = 0

⇒ x2 – 60x + 48x – 2880 = 0

⇒ x(x – 60) + 48(x – 60) = 0

⇒ (x – 60)(x + 48) = 0

⇒ x = 60 or x = - 48

⇒ x = 60 (Rejecting x = - 48, as speed can’t be negative)

Hence, speed of the express train = 60 km/h

27.A car covers a distance of 400 km at a certain speed. Has the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.

Let the original speed of the car = x km/h

Distance covered = 400 km

Time taken to cover 400 km = 400/x h

In second case,

Speed of car, (x + 12) km/h

New time taken to cover 400 km = 400/(x + 12)h

According to the condition

400x – 400/(x + 12)

= 1.40/60

= 1.2/3

= 5/3

⇒ 400(x + 12 – x)/{x(x + 12)} = 5/3

⇒ (400 × 12)/(x2 + 12x) = 5/3

400 × 12 × 3 = 5x2 + 60x

⇒ 14400 = 5x2 + 60x

⇒ 5x2 + 60x – 14400 = 0

⇒ x2 + 12x – 2880 = 0 (dividing both side by 5)

⇒ x2 + 60x – 48x – 2880 = 0

⇒ x(x + 60) – 48(x + 60) = 0

⇒ (x + 60)(x – 48) = 0

⇒ x = 48 or x = - 60

⇒ x = 48 (Rejecting x = - 60, being speed)

Hence, original speed of the car = 48 km/h

28.An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for

(i) the onward journey,

(ii) the return journey,

If the return journey took 30 minutes less than the onward journey, write down an equation in x find its value.

Distance = 400 km

Speed of aeroplane = x km/hr

(i) ∴ Time taken = 400/x hours

On increasing the speed by 40 km/hr,

On the return journey, the speed = (x + 40) km/hr.

(ii) Time taken = 400/(x + 40) hours

Now according to the condition,

400/x – 400/(x + 40) = 30 minutes

= 1/2

400 [(1/x) – 1/(x + 40)] = 1/2

⇒ (400 × 40)/(x2 + 40x) = 1/2

⇒ x2 + 40x = 400 × 40 × 2

⇒ x2 + 40x – 32000 = 0

⇒ x2 + 200x – 160x – 32000 = 0

⇒ x(x + 200) – 160(x + 200) = 0

⇒ (x + 200)(x – 160) = 0

Either x + 200 = 0, then x = - 200, which is not possible as it is negative.

or x – 160 = 0, then x = 160

29. The distance by road between two towns A and B, is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr, and the train travels at a speed which is 16 km/hr faster than the car. Calculate:

(i) The time taken by the car, to reach town B from A, in terms of x;

(ii) The time taken by the train, to reach town B from A, in terms of x;

(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.

(iv) Hence find the speed of the train.

The distance by road between A and B = 216 km

And the distance by rail = 208 km

Speed of car = x km/hr

And speed of train = (x + 16) km/hr

(i) Time taken by car = 216/x hours

(ii) Time taken by car = 208/(x + 16) hours

(iii) According to the condition,

216/x – 208/(x + 16) = 2

⇒ (216x + 216 × 16 – 208x)/{x(x + 16)} = 2/1

⇒ (8x + 3456)/(x2 + 16x) = 2/1

⇒ 8x + 3456 = 2x2 + 32x

⇒ 2x2 + 32x – 8x – 3456 = 0

⇒ 2x2 + 24x – 3456 = 0

⇒ x2 + 12x – 1728 = 0 (Dividing by 2)

⇒ x2 + 48x – 36x – 1728 = 0

⇒ x(x + 48) – 36(x + 48) = 0

⇒ (x + 48)(x – 36) = 0

Either x + 48 = 0, then x = - 48, which is not possible as it is negative.

Or x – 36 = 0, then x = 36

(iv) Speed of the train = (x + 16) km/hr.

= (36 + 16)km/hr

= 52 km/hr

30. An aeroplane flying with a wind of 30 km/hr takes 40 minutes less to fly 3600 km, that what it would have taken to fly against the same wind. Find the planes speed of flying in still air.

Let the speed of the plane in still air = x km/hr

Speed of wind = 30 km/hr

Distance = 3600 km

∴ Time taken with the wind = 3600/(x + 30)

And time taken against the wind = 3600/(x – 30)

According to the condition,

3600/(x – 30) – 3600/(x + 30)

= 40 minutes

= 2/3 hour

⇒ (3600 × 60)/(x2 – 900) = 2/3

⇒ 2x2 – 1800

= 3 × 3600 × 60

⇒ 2x2 – 1800 = 64800

⇒ 2x2 – 1800 – 648000 = 0

⇒ 2x2 – 649800 = 0

⇒ x2 – 324900 = 0 (Dividing by 2)

⇒ x2 - (570)2 = 0

⇒ (x + 570)(x – 570) = 0

Either x + 570 = 0, then x = - 570 which is not possible as it is negative

Or x – 570 = 0, then x = 570

Hence speed of plane in still air = 570 km/hr

31. A school bus transported an excursion party to a picnic spot 150 km away. While returning, it was raining and the bus had to reduce its speed by 5 km/hr, and it took one hour longer to make the return trip. Find the time taken to return.

Distance = 150 km

Let the speed of bus = x km/hr

∴ Time taken = 150/x hour

On returning speed of the bus = (x – 5) km/hr

∴ Time taken = 150/(x – 5)

According to the condition

150/(x – 5) – 150/x) = 1

⇒ 150{1/(x – 5) – 1/x} = 1

⇒ 150[(x – x + 5)/{x(x – 5)}] = 1

⇒ (150 × 5)/(x2 – 5x) = 1

⇒ x2 – 5x = 750

⇒ x2 – 5x – 750 = 0

⇒ x2 – 30x + 25x – 750 = 0

⇒ x(x – 30) + 25(x – 30) = 0

⇒ (x – 30)(x + 25) = 0

Either x – 30 = 0, then x = 30

Or x + 25 = 0, then x = - 25, but it is not possible as it is negative.

∴ Speed of bus = 30 km and time taken while returning

= 150/(x – 5)

= 150/(30 – 5)

= 150/25

= 6 hours

32. A boat can cover 10 km up the stream and 5 km down the stream in 6 hours. If the speed of the stream is 1.5 km/hr. find the speed of the boat in still water.

Distance up stream = 10 km

And down stream = 5 km

Total time is taken = 6 hours

Speed of stream = 1.5 km/hr

Let the speed of a boat in still water = x km/hr

According to the condition,

10/(x – 1.5) + 5/(x + 1.5) = 6

⇒ 15x + 7.5 = 6 (x2 – 2.25)

⇒ 15x + 7.5 = 6x2 – 13.5

⇒ 6x2 – 15x – 13.5 – 7.5

⇒ 6x2 – 15x – 21 = 0

⇒ 2x2 – 5x – 7 = 0 (Dividing by 3)

⇒ 2x2 – 7x + 2x – 7 = 0

{2 × (-7) = - 14, - 14 = - 7 × 2 , - 5 = - 7 + 2

⇒ {x(2x – 7)} + 1(2x – 7) = 0

⇒ (2x – 7)(x + 1) = 0

Either 2x – 7 = 0 then 2x = 7

⇒ x = 7/2

Or x + 1 = 0, then x = - 1

But it is not possible being negative.

∴ x = 7/3 = 3.5

∴ Speed of boat = 3.5 km/hr

33. Two pipes running together can fill a tank in 111/9 minutes. If one pipe takes 5 minutes more than the other to fill the tank, find the time in which each pipe would/fill the tank.

Let the time taken by one pipe = x minutes

Then time taken by second pipe = (x + 5) minutes

Time taken by both pipes = 111/9 minutes

Now according to the condition,

1/x + 1/(x + 5) = 9/100

⇒ {(x + 5) + x}/{x(x + 5) = 9/100

⇒ (x + 5 + x)/(x2 + 5x) = 9/100

⇒ (2x + 5)/(x2 + 5x) = 9/100

⇒ 9x2 + 45x = 200x + 500

⇒ 9x2 + 45x – 200x – 500 = 0

⇒ 9x2 – 155x – 500 = 0

⇒ 9x2 – 180x + 25x – 500 = 0

⇒ 9x(x – 20) + 25(x – 20) = 0

⇒ (x – 20)(9x + 25) = 0

Either x – 20 = 0, then x = 20

Or 9x + 25 = 0, then 9x = - 25

⇒ x = - 25/9 but it is not possible as it is negative

∴ x = 20

Hence the first pipe can fill the tank in 20 minutes and second pipe do the same in 20 + 5

= 25 mints.

34. (i) Rs 480 is divided equally ‘x’ children. If the number of children was 20 then each would have got Rs 12 less. Find ‘x’.

(ii) Rs 6500 is divided equally among a certain number of persons. Had there been 15 more persons, each would have got Rs. 30 less. Find the original number of persons.

(i) Share of each child = Rs 480/x

According to the question,

480/(x + 20) = 480/x – 12

⇒ 480/(x + 20) = (480 – 12x)/x

⇒ 480/(x + 20) = {12(40 – x)}/x

⇒ (x + 20)(40 – x) = 40x

⇒ 40x – x2 + 800 – 20x = 40x

⇒ x2 + 20x – 800 = 0

⇒ x2 + 40x – 20x – 800 = 0

⇒ x(x + 40) – 20(x + 40) = 0

⇒ (x + 40)(x – 20) = 0

⇒ x = - 40, x = 20

- ve value of x is not possible

∴ No. of children = 20

(ii) Total amount = Rs 6500

Let number of persons in first case = x

Then share of each person = Rs 6500/x

Number of persons increased = 15

∴ Total number = x + 15

Now share of each person = 6500/(x + 15)

According to the condition,

6500/x – 6500/(x + 15) = 30

⇒ 6500[1/x – 1/(x + 15)] = 30

⇒ 6500[(x + 15 – x)/{x(x + 15)}] = 30

⇒ (6500 × 15)/(x2 + 15x) = 30

⇒ 6500 × 15 = 30(x2 + 15x)

⇒ 3250 = x2 + 15x (Dividing by 30)

⇒ x2 + 15x – 3250 = 0

⇒ x2 + 65x – 50x – 3250 = 0

⇒ x(x + 65) – 50(x + 65) = 0

⇒ (x + 65)(x – 50) = 0

Either x + 65 = 0, then x = - 65 which is not possible being negative

Or x – 50 =0, then x = 50

∴ No. of persons in the beginning = 50

35. 2x articles cost Rs(5x + 54) and (x + 2) similar articles cost Rs(10x – 4), find x.

Cost of 2x articles = 5x + 54

Cost of 1 article = (5x + 54)/2x …(i)

Again cost of x + 2 articles = 10x – 4

∴ Cost of 1 article = (10x – 4)/(x + 2) ….(ii)

From (i) and (ii),

(5x + 54)/2x = (10x – 4)/(x + 2)

⇒ (5x + 54)(x + 2) = 2x(10x – 4)

⇒ 5x2 + 10x + 54x + 108 = 20x2 – 8x

⇒ 5x2 – 10x + 54x + 108 – 20x2 + 8x = 0

⇒ - 15x2 + 72x + 108 = 0

⇒ 5x2 – 24x – 36 = 0 (Dividing by – 3)

⇒ 5x2 – 30x + 6x – 36 = 0

⇒ 5x(x – 6) + 6(x – 6) = 0

⇒ (x – 6)(5x + 6) = 0

Either x – 6 = 0, then x = 6

Or 5x + 6 = 0, then 5x = - 6

⇒ x = -6/5, but it is possible as it is in negative.

∴ x = 6

36. A trader buys x articles for a total cost of Rs 600.

(i) Write down the cost of one article in terms of x. If the cost per article were Rs 5 more, the number of articles than can be bought for Rs 600 would be four less.

(ii) Write down the equation in x for the above situation and solve it to find x.

Total cost = Rs 600,

No. of articles = x

(i) ∴ Cost of one article = Rs 600/x

In second case price of one article = Rs 600/x + 5

= 600/(x – 4)

(ii) (600 + 5x)/x = 600/(x – 40

⇒ (x – 4)(600 + 5x) = 600x

⇒ 600x – 2400 + 5x2 – 20x = 600x

⇒ 5x2 + 600x – 20x – 600x – 2400 = 0

⇒ 5x2 – 20x – 2400 = 0

⇒ x2 – 4x – 480 = 0 (Dividing by 5)

⇒ x2 – 24x + 20x – 480 = 0

⇒ x(x – 24) + 20(x – 24) = 0

⇒ (x – 24)(x + 20) = 0

Either x – 24 = 0, then x = 24

Or x + 20 = 0, then x = - 20, but it is not possible as it is negative.
∴ x = 24

37. A shopkeeper buys a certain number of books for Rs 960. If the cost per book was Rs 8 less, the number of books that could be bought for Rs 960 would be 4 more. Taking the original cost of each book to be Rs x, write an equation in x and solve it to find the original cost of each book.

Let original cost = Rs x
No. of books bought = 960/x
New cost of books = Rs(x – 8)
∴ No. of books bought = 960/(x – 8)
If no. books bought in 4 more then cost = (960/x) + 4
∴ According to condition,
960/(x – 8) – 960/x = 4
⇒ 960 {1/(x – 8) – 1/x} = 4
⇒ {x – (x – 8)}/{x(x – 8)} = 4/960
⇒ (x – x + 8)/(x2 – 8x) = 4/960
⇒ 8/(x2 – 8x) = 1/240
⇒ x2 – 8x = 8 × 240
⇒ x2 – 8x – 1920 = 0 = (8 ± 88)/2
= (8 + 88)/2, (8 – 88)/2
= 96/2, (-80)/2
= 48, - 40 (rejecting)
∴ Cost of book = ₹ 48.

38. A piece of cloth Rs 300. If the piece was 5 metres longer and each metre of cloth costs Rs 2 less, the cost of the piece would have remained unchanged. How long is the original piece of cloth and what is the rate per metre ?

The total cost of cloth piece = Rs 300
Let the length of the piece of cloth in the beginning = x m
Then cost of 1 metre = Rs 300/x
In second case, length of cloth = (x + 5)
Cost of 1 metre = Rs 300/(x + 5)
According to the condition,
300/x – 300/(x + 5) = 2
⇒ 300(1/x – 1/(x + 5) = 2
⇒ 300[(x + 5 – x)/{x(x + 5)}] = 2
⇒ (300 × 5)/{x(x + 5)} = 2
⇒ (150 × 5)/{x(x + 5) = 1 (Dividing by 2)
⇒ 750 = x2 + 5x
⇒ x2 + 5x – 750 = 0
⇒ x2 + 30x – 25x – 750 = 0
⇒ x(x + 30) – 25(x + 30) = 0
⇒ (x + 30)(x – 25) = 0
Either x + 30 = 0, then x = - 30 which is not possible being negative
or x – 25 = 0, then x = 25
∴ Length of cloth piece in the beginning = 25 metres
And rate per metre = Rs 300/25 = Rs 12

39. The hotel bill for a number of people for an overnight stat is Rs 4800. If there were 4 more, the bill each person had to pay would have reduced by Rs 200. Find the number of people staying overnight.

Let the number of people = x
Amount of bill = Rs 4800
Then bill for each person = Rs 4800/x
In second case,
The number of people = x + 4
Then bill of each person = 4800/(x + 4)
According to the condition,
4800/x – 4800/(x + 4) = 200
⇒ 4800(1/x – 1/(x + 4) = 200
⇒ 4800(x + 4 – x){x(x + 4)} = 200
⇒ (4800 × 4)/{x(x + 4)} = 200
⇒ 19200 = 200x2 + 800x
⇒ 200x2 + 800x – 19200 = 0
⇒ x2 + 4x – 96 = 0 (Dividing by 200)
⇒ x2 + 12x - 8x – 96 = 0
⇒ x(x + 12) – 8(x + 12) = 0
⇒ (x + 12)(x – 8) = 0
Either x + 12 = 0, then x = - 12, but it is not possible as it is in negative.
Or x – 8 = 0, then x = 8
∴ No. of people = 8

40. A person was given Rs 3000 for a tour. If he extends his tour programme by 5 days, he must cut down his daily expenses by Rs 20. Find the number of days of his tour programme.

Let the number of days of tour programme = x

Amount = Rs 3000

∴ Expenses for each day = 3000/x

In second case, no. of days = x + 5

Then expenses of each day = 3000/(x + 5)

Now according to the condition,

3000/x – 3000/(x + 5) = 20

⇒ 3000[1/x – {1/(x + 5)}] = 20

⇒ 3000{(x + 5 – x)/(x2 + 5x)} = 20

⇒ 3000 × 5 = 20x2 + 100x

⇒ 20x2 + 100x – 15000 = 0

⇒ x2 + 5x – 750 = 0 (Dividing by 20)

⇒ x2 – 25x + 30x – 750 = 0

⇒ x(x – 25) + 30(x – 25) = 0

⇒ (x – 25)(x + 30) = 0

Either x – 25 = 0, then x = 25

Or x + 30 = 0, then x = - 30, but it is not possible as it is in negative.

∴ Number days = 25

41. Ritu bought a saree for Rs 60x and sold it for Rs (500 + 4x) at a loss of x%. Find the cost price.

The cost price of saree = Rs 60x

And selling price = Rs(500 + 4x)

Loss = x%

Now according to the condition

S.P. = C.P. × (100 – Loss %)/100

500 + 4x = {60x(100 – x)}/100

⇒ 50000 + 400x = 600x – 60x2

⇒ 60x2 – 600x + 400x + 50000 = 0

⇒ 60x2 – 5600 + 50000 = 0

⇒ 3x2 – 280x + 2500 = 0 (Dividing by 20)

⇒ 3x2 – 30x – 250x + 2500 = 0

⇒ 3x(x – 10) – 250(x – 10) = 0

⇒ (x – 10)(3x – 250) = 0

Either x – 10 = 0, then x = 10

Or 3x – 250 = 0, then 3x = 250

⇒ x = 250/3

But it is not possible

∴ Loss = 10%

Cost price = 60x = 60 × 10

= Rs 600

42. (i) The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages.

(ii) Paul is x years old and his father’s age is twice the square of Paul’s age. Ten years hence, the father’s age will be four times Paul’s age. Find their present ages.

(i) Let Vivek’s present age be x years.

His brother age = (47 – x) years

According to question,

x(47 – x) = 550

⇒ 47x – x2 = 550

⇒ x2 – 47x + 550 = 0

⇒ x2 – 25x – 22x + 550 = 0

⇒ x(x – 25) – 22(x – 25) = 0

⇒ (x – 25) (x – 22) = 0

⇒ x – 25 = 0 or x – 22 = 0

⇒ x = 25 or x = 22

When x = 25, then 47 – x = 47 – 25 = 22

When x = 22, then 47 – x = 47 – 22 = 25 (does not satisfy the given condition)

∴ Vivek’s age = x = 25 years

His younger brother’s age = 22 years

(ii) Age of Paul = x years

Father’s age = 2x2

10 years hence,

Age of Paul = x + 10

And Father’s age = 2x2 + 10

According to the conditions,

2x2 + 10 = 4(x + 10)

⇒ 2x2 + 10 = 4x + 40

⇒ 2x2 + 10 – 4x – 40 = 0

⇒ 2x2 – 4x – 30 = 0

⇒ x2 – 2x – 15 = 0 (Dividing by 2)

⇒ x2 – 5x + 3x – 15 = 0

⇒ x(x – 5) + 3(x – 5) = 0

⇒ (x – 5)(x + 3) = 0

Either x – 5 = 0, then x = 5

Or x + 3 = 0, then x = - 3, but it is not possible as it is in negative.

∴ Age of Paul = 5 years and his father’s age = 2x2 = 2(5)2

= 2 × 25

= 50 years

43. The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than three times the age of his son. Find the present age.

Let the present age of the son = x years

Then, the present age of the man = 2x2 years

8 years hence,

The age of son will be = (x + 8) years and the

Age of man = (2x2 + 8) years

According to the problem,

2x2 + 8 = 3(x + 8)

⇒ 2x2 + 8 = 3x + 24 + 4

⇒ 2x2 – 3x – 24 – 4 + 8 = 0

⇒ 2x2 – 3x – 20 = 0

⇒ 2x2 – 8x + 5x – 20 = 0

⇒ 2x(x – 4) + 5(x – 4) = 0

⇒ (x – 4)(2x + 5) = 0

Either x – 4 = 0, then x = 4

Or 2x + 5 = 0 then 2x = - 5

⇒ x = - (5/2)

But, it is not possible.

Present age of the son = 4 years

And present age of the man = 2x2

= 2(4)2 years

= 32 years

44. Two years ago, a man’s age was three times the square of his daughter’s age. Three years hence, his age will be four times his daughter’s age. Find their ages.

2 years ago,

Let the age of daughter = x

Age of man = 3x2

Then present age of daughter = x + 2

And mean = 3x2 + 2

And 3 years hence, the age of the daughter = x + 2 + 3

= x + 5

and man = 3x2 + 2 + 3 = 3x2 + 5

According to the condition,

3x2 + 5 = 4(x + 5)

⇒ 3x2 + 5 = 4x + 20

⇒ 3x2 – 4x + 5 – 20 = 0

⇒ 3x2 – 4x – 15 = 0

⇒ 3x2 – 9x + 5x – 15 = 0

⇒ 3x(x – 3) + 5(x – 3) = 0

⇒ (x – 3)(3x + 5) = 0

Either x – 3 = 0, then x = 3

Or 3x + 5 = 0, then 3x = - 5

⇒ x = - 5/3 which is not possible, as age can’t be negative

If x = 3, then

Present age of man = 3x2 + 2

= 3(3)2 + 2

= 27 + 2

= 29 years

And age of daughter = x + 2

= 3 + 2

= 5 years

45. The length (in cm) of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of another side by 1 cm. Find the length of each side. Also, find the perimeter and the area of the triangle.

Let the length of one side = x cm

And other side = y cm

Then hypotenuse = x + 2, and 2y + 1

∴ x + 2 = 2y + 1

⇒ x – 2y = 1 – 2

⇒ x = 2y – 1 ...(i)

And by using Pythagoras theorem,

x2 + y2 = (2y + 1)2

⇒ x2 + y2 = 4y2 + 4y + 1

⇒ (2y – 1)2 + y2 = 4y2 + 4y + 1 [From (i)]

⇒ 4y2 – 4y + 1 + y2 = 4y2 + 4y + 1

⇒ 4y2 – 4y + 1 + y2 – 4y2 – 4y – 1 = 0

⇒ y2 – 8y = 0

⇒ y(y – 8) = 0

Either y = 0, but it is not possible.

Or y – 8 = 0 then y = 8

Substituting the value of y in (i)

x = 2(8) – 1

= 16 – 1

= 15

∴ Length of one side = 15 cm

and length of other side = 8 cm

and hypotenuse = x + 2

= 15 + 2

= 17

∴ Perimeter = 15 + 8 + 17

= 40 cm

And area = 1/2 × one side × other side

= 1/2 × 15 × 8

46. If twice the area of a smaller is subtracted from the area of a larger square, the result is 14 cm2. However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm2. Determine the sides of the two squares.

Let the side of smaller square = x cm

And side of bigger square = y cm

According to the condition,

y2 – 2x2 = 14 ….(i)

and 2y2 + 3x2 = 203 ...(ii)

Multiply (i) by 2 and (ii) by 1 ⇒ x2 = (- 175)/(- 7)

= 25

x2 – 25 = 0

⇒ (x + 5)(x – 5) = 0

Either x + 5 = 0 then x = - 5, but it is not possible, or x – 5 = 0, then x = 5

Substitute the value of x in (i)

y2 – 2(5)2 = 14

⇒ y2 = 14 + 2 × 25

y2 = 14 + 50 = 64

= (8)2

∴ y = 8

Hence side of smaller square = 5 cm

And side of bigger square = 8 cm

### Multiple Choice Type

Choose the correct answer from the given four options:

1. Which of the following is not a quadratic equation ?

(a) (x + 2)2 = 2(x + 3)

(b) x2 + 3x = (- 1) (1 – 3x)

(c) (x + 2)(x – 1) = x2 – 2x – 3

(d) x3 – x2 + 2x + 1 = (x + 1)3

(c) (x + 2)(x – 1) = x2 – 2x – 3

(a) (x + 2)2 = 2(x + 3)

⇒ x2 + 4x + 4 = 2x + 6

⇒ x2 + 4x – 2x + 4 – 6 = 0

⇒ x2 + 2x – 2

(b) x2 + 3x

= (-1)(1 – 3x)

⇒ x2 + 3x = - 1 + 3x

⇒ x2 + 1 = 0

(c) (x + 2)(x – 1) = x2 – 2x – 3

x2 – x + 2x – 2 = x2 – 2x – 3

x2 – x2 + x + 2x – 2 + 3 = 0

⇒ 3x + 1 = 0

It is not a quadratic equation.

(d) x3 – x2 + 2x + 1 = (x + 1)33

= x3 + 3x2 + 3x + 1

x3 – x2 + 2x + 1

3x2 + x2 – 2x – 1 + 3x + 1 = 0

⇒ 4x2 + x = 0

2. Which of the following is a quadratic equation ?

(a) (x – 2)(x + 1) = (x – 1)(x – 3)

(b) (x + 2)3 = 2x(x2 – 1)

(c) x2 + 3x + 1 = (x – 2)2

(d) 8(x – 2)3 = (2x – 1)3 + 3

(d) 8(x – 2)3 = (2x – 1)3 + 3

(a) (x – 2)(x + 1) = (x – 1)(x – 3)

⇒ x2 + x – 2x – 2 = x2 – 3x – x + 3

⇒ 3x + x – 2x + x = 3 + 2

⇒ 3x = 5

It is not a quadratic equation.

(b) (x + 2)3 = 2x(x2 – 1)

x3 + 6x2 + 12x + 8 = 2x3 – 2x

x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0

- x3 + 6x2 + 14x + 8 = 0

It is not a quadratic equation.

(c) x2 + 3x + 1 = (x – 2)2

x2 + 3x + 1 = x2 – 4x + 4

3x + 1 + 4x – 4 = 0

⇒ 7x – 3 = 0

It is not a quadratic equation.

(d) 8(x – 2)3 = (2x – 1)3 + 3

8(x3 – 6x2 + 12x – 8)

= 8x3 – 12x2 + 6x – 1 + 3

8x3 – 48x2 + 96x – 64 – 8x3 + 12x2 – 6x + 1 – 3 = 0

- 36x2 + 90x – 66 = 0

3. Which of the following equations has 2 as a root ?

(a) x2 – 4x + 5 = 0

(b) x2 + 3x – 12 = 0

(c) 2x2 – 7x + 6 = 0

(d) 3x2 – 6x – 2 = 0

(c) 2x2 – 7x + 6 = 0

(a) x2 – 4x + 5 = 0

⇒ (2)2 – 4x2 + 5 = 0

⇒ 4 – 8 + 5 = 0

⇒ 9 – 8 ≠ 0

2 is not its root.

(b) x2 + 3x – 12 = 0

⇒ (2)2 – 3 × 2 – 12 = 0

⇒ 4 – 6 – 12

= 4 – 18

= - 14

∴ 2 is not its roots.

(c) 2x2 – 7x + 6 = 0

⇒ 2(2)2 – 7 × 2 + 6 = 0

⇒ 8 – 14 + 6 = 0

⇒ 0 = 0

∴ 2 is its root.

(d) 3x2 – 6x – 2 = 0

⇒ 3(2)2 – 6 × 2 – 2 = 0

⇒ 12 – 12 – 2 = 0

⇒ 12 – 14 = 0

∴ 2 is not its root.

4. If 1/2 is a root of the equation x2 + kx- 5/4 = 0, then the value of k is

(a) 2

(b) – 2

(c) 1/4

(d) 1/2

(a) 2

1/2 is a root of the equation

x2 + kx – 5/4 = 0

Substituting the value of x = 1/2 in the equation

(1/2)2 + k × 1/2 – 5/4 = 0

⇒ 1/4 + k/2 – 5/4 = 0

⇒ k/2 – 1 = 0

⇒ k = 1 × 2 = 2

∴ k = 2

5. If 1/2 is a root of the quadratic equation 4x2 – 4kx + k + 5 = 0, then the value of k is

(a) – 6

(b) – 3

(c) 3

(d) 6

1/2 is a rot of the equation

4x2 – 4kx + k + 5 = 0

Substituting the value of x = 1/2 in the equation

4(1/2)2 – 4 × k × 1/2 + k + 5 = 0

1 – 2k + k + 5 = 0

⇒ - k + 6 = 0

k = 6

6.The roots of the equation x2 – 3x – 10 = 0 are

(a) 2, - 5

(b) – 2, 5

(c) 2, 5

(d) – 2, - 5

(b) – 2, 5

x2 – 3x – 10 = 0 = (3 + 7)/2

∴ x = (3 + 7)/2 = 5 and x = (3 – 7)/2 = (-4)/2 = - 2

x = 5, - 2 or – 2, 5

7. If one root of a quadratic equation with rational coefficients is (3 - √5)/2, then the other

(a) (– 3 - √5)/2

(b) (– 3 + √5)/2

(c) (3 + √5)/2

(d) (√3 + 5)/2

(c) (3 + √5)/2

One root of a quadratic equation is (3 - √5)/2 then other root will be (3 + √5)/2

8. If the equation 2x2 + 5x + (k + 3) = 0

Has equal roots then the value of k is

(a) 9/8

(b) – (9/8)

(c) 1/8

(d) – (1/8)

(c) 1/8

2x2 – 5x + (k + 3) = 0

a = 2, b = - 5, c = k + 3

∴ b2 – 4ac = (5)2 – 4 × 2 × (k + 3)

= 25 – 8(k + 3)

∴ Roots are equal.

∴ b2 – 4ac = 0

∴ 25 – 8(k + 3) = 0

25 – 8k – 24 = 0

1 – 8k = 0

∴ 8k = 1

∴ k = 1/8

9. The value(s) of k which the quadratic equation 2x2 – kx + k = 0 has equal roots is (are)

(a) 0 only

(b) 4

(c) 8 only

(d) 0, 8

(d) 0, 8

2x2 – kx + k = 0

a = 2, b = - k, c = k

∴ b2 – 4ac = (-k)2 – 4 × 2 × k

= k2  - 8k = 0

∴ Roots are equal.

∴  b2 – 4ac = 0

k2 – 8k = 0

⇒ k(k – 8) = 0

Either k = 0

Or k – 8 = 0, then k = 8

k = 0, 8

10. If the equation 3x2 – kx + 2k = 0 roots, then the value(s) of k is are

(a) 6

(b) 0 only

(c) 24 only

(d) 0

(d) 0

3x2 – kx + 2k = 0

Here, a= 3, b = - k, c = 2k

b2 – 4ac

= (- k)2 – 4 × 3 × 2k

= k2 – 24k

∴ Roots are equal.

∴ b2 – 4ac = 0

∴  k2 – 24k = 0

∴  k(k – 24) = 0

Either k = 0,

Or, k – 24 = 0, then k = 24

∴  k = 0, 24

11. If the equation {(k + 1)x2 – 2(k – 1)x + 1 = 0 has equal roots, then the values of k are

(a) 1, 3

(b) 0, 3

(c) 0, 1

(d) 0, 1

(b) 0, 3

(k + 1)x2  – 2(k – 1)x + 1 = 0

Here,  a = k + 1, b = - (k – 1),  c = 1

∴ b2 – 4ac

= [-2(k – 1)2 – 4(k + 1)(1)

= 4(k2 – 2k + 1) – 4k – 4

= 4k2 – 8k + 4 – 4k – 4

= 4k2 – 12k

∵ Roots are equal.

∴ b2 – 4ac = 0

∵ 4k2 – 12k = 0

⇒ 4k(k – 3) = 0

⇒ k(k – 3) = 0

Either k = 0

Or k – 3 = 0, then k = 3

12. If the equation 2x2 – 6x + p = 0 has real and different roots, then the values of p are given by

(a) p < 9/2

(b) p 9/2

(c) p > 9/2

(d) p 9/2

(a) p < 9/2

2x2 – 6x + p = 0

Here, a = 2, b = - 6, c = p

b2 – 4ac

= (-6)2 – 4 × 2 × p

= 36 – 8p

∵  Roots are real and unequal.

∴  b2 – 4ac > 0

⇒ 36 – 8p > 0

⇒ 36 > 8p

⇒ 36/8 > p

⇒ p < 36/8

⇒ p < 9/2

13. The quadratic equation 2x2 - √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than two real roots

(c) no real roots

2x2 - √5x + 1 = 0

Here, a = 2, b = - √5, c = 1

b2 – 4ac

= (-5)2 – 4 × 2 × 1

= 5 – 8

= - 3

∵ b2 – 4ac < 0

∴ It has no real roots.

14. Which of the following equations has two distinct real roots ?

(a) 2x2 - 3√2x + 9/4 = 0

(b) x2 + x – 5 = 0

(c) x2 + 3x + 2√2 = 0

(d) 5x2 – 3x + 1 = 0

(a) 2x2 - 3√2x + 9/4 = 0

b2 – 4ac

= (1)2 – 4 × 1 × (-5)

= 1 + 20

= > 0

Roots are real and distinct.

15. Which of the following equations has no real roots?

(a) x2 – 4x + 3√2 = 0

(b) x2 + 4x - 3√2 = 0

(c) x2 – 4x - 3√2 = 0

(d) 3x2 + 4√3x + 4 = 0

(a) x2 – 4x + 3√2 = 0

b2 – 4ac

= (4)2 – 4 × 1 × 3√2

or 16 - 12√2

so 16 – 12(1.4)

Therefore, 16 – 16.8

Hence - 0.8

b2 – 4ac < 0

Roots are not real. (a)

### Chapter Test

Solve the following equations (1 to 4) by factorisation:

1. (i) x2 + 6x – 16 = 0

(ii) 3x2 + 11x + 10 = 0

(i) x2 + 6x – 16 = 0

⇒ x2 + 8x – 2x – 16 = 0

⇒ x(x + 8) – 2(x + 8) = 0

⇒ (x + 8)(x – 2) = 0

Either x + 8 = 0, then x = - 8

Or x – 2 = 0, then x = 2

Hence x = - 8, 2

(ii) 3x2 + 11x + 10 = 0

3x2 + 11x + 10 = 0

⇒ 3x2 + 6x + 5x + 10 = 0

3x(x + 2) + 5(x + 2) = 0

⇒ (x + 2)(3x + 5) = 0

Either x + 2 = 0, then x = - 2

Or 3x + 5 = 0, then

3x = - 5

⇒ x = -5/3

Hence, x = - 2, -5/3

2. (i) 2x2 + ax – a2 = 0

(ii) √3x2 + 10x + 7√3 = 0

(i) 2x2 + ax – a2 = 0

⇒ 2x2 + 2ax – ax – a2 = 0

⇒ 2x(x + a) – a(x + a) = 0

⇒ (x + a)(2x – a) = 0

Either x + a = 0, then x = - a

or 2x – a = 0, then

2x = a

⇒ x = a/2

Hence = x = - a, a/2

(ii) √3x2 + 10x + 7√3 = 0

√3x2 + 10x + 7√3 = 0

⇒ √3x2 + 3x + 7x + 7√3 = 0

⇒ √3x(x + √3) + 7(x + √3) = 0

⇒ (x + √3)( √3x + 7) = 0

Either x + √3 = 0, then x = - √3

or √3x + 7 = 0, then √3x = - 7

⇒ x = -7/√3

x = (- 7 ×√3)/(√3 ×√3)

= (-7√3)/3

Hence, x = - √3, (-7√3)/3

3. (i) x(x + 1) + (x + 2)(x + 3) = 42

(ii) 6/x – 2/(x – 1) = 1/(x – 2)

(i) x(x + 1) + (x + 2)(x + 3) = 42

⇒ 2x2 + 6x + 6 – 42 = 0

⇒ x2 + 3x – 18 = 0 (Dividing by 2)

⇒ x2 + 6x – 3x – 18 = 0

⇒ x(x + 6) – 3(x + 6) = 0

⇒ (x + 6)(x – 3) = 0

Either x + 6 = 0, then x = - 6

Or x – 3 = 0, then x = 3

Hence x = - 6, 3

(ii) 6/x – 2/(x – 1) = 1/(x – 2)

⇒ (6x – 6 – 2x)/{x(x – 1)} = 1/(x – 2)

⇒ (4x – 6)/(x2 – x) = 1/(x – 2)

⇒ 4x2 – 8x – 6x + 12 = x2 – x

⇒ 4x2 – 14x + 12 – x2 + x = 0

⇒ 3x2 – 13x + 12 = 0

⇒ 3x2 – 9x – 4x + 12 = 0

⇒ 3x(x – 3) – 4(x – 3) = 0

⇒ (x – 3)(3x – 4) = 0

Either x – 3 = 0, then x = 3

or 3x – 4 = 0, then

3x = 4 ⇒ x = 4/3

Hence, x = 3, 4/3 Squaring on both sides

x + 15 = (x + 3)2

⇒ x + 15 = x2 + 6x + 9

⇒ x2 + 6x + 9 – x – 15 = 0

⇒ x2 – 5x – 6 = 0

⇒ x2 + 6x – x – 6 = 0

⇒ x(x + 6) – 1(x + 6) = 0

⇒ (x + 6)(x – 1) = 0

Either x + 6 = 0, then x = - 6

or x – 1 = 0, then x = 1

∴ x = - 6, 1

Check:

If x = - 6 then = 3

R.H.S. = x + 3

= - 6 + 3

= - 3

∵ L.H.S. = R.H.S.

∴ x = 1 is a root of this equation

Hence x = 1 Squaring both sides

3x2 – 2x – 1 = (2x – 2)2

⇒ 3x2 – 2x – 1 = 4x2 – 8x + 4

⇒ 4x2 – 8x + 4 – 3x2 + 2x + 1 = 0

⇒ x2 – 6x + 5 = 0

⇒ x2 – 5x – x + 5 = 0

⇒ x(x – 5) – 1(x – 5) = 0

⇒ (x – 5)(x – 1) = 0

Either x – 5 = 0, then x = 5

Or x – 1 = 0, then x = 1

Check:

If x = 5, then = 8

R.H.S. = 2x – 2

= 2 × 5 – 2

= 10 – 2

= 8

∵ L.H.S = R.H.S.

∴ x = 5 is a root.

If x = 1, then

R.H.S. = 2x – 2

= 2 × 1 – 2

= 2 – 2 = 0

∵ L.H.S = R.H.S.

∴ x = 1 is also its root.

Hence, x = 5, 1

5. (i) 2x2 – 3x – 1 = 0

(ii) x (3x + 1/2) = 6

(i) 2x2 – 3x – 1 = 0

Here a = 2, b = - 3, c = - 1

D = b2 – 4ac

= (3)2 – 4 × 2 × (-1)

= 9 + 8

= 17

∵ x = (- b ± D)/2a (ii) x(3x + 1/2) = 6

3x2 + x/2 = 6

⇒ 6x2 + x = 12

⇒ 6x2 + x – 12 = 0

Here a = 6, b = 1, c = - 12

D = b2 – 4ac

= (1)2 – 4 × 6 × (-12)

= 1 + 288

= 289 = (- 1 ± 17)/12

∴ x1 = (- 1 + 17)/12 = 16/12 = 4/3

x2 = (- 1 – 17)/12

= - 18/12

= - (3/2)

∴ x = 4/3, - (3/2)

6. (i) (2x + 5)/(3x + 4) = (x + 1)/(x + 3)

(ii) 2/(x + 2) – 1/(x + 1) = 4/(x + 4) – 3/(x + 3)

(i) (2x + 5)/(3x + 4) = (x + 1)(x + 3)

(2x + 5)(x + 3) = (x + 1)(3x + 4)

2x2 + 6x + 5x + 15 = 3x2 + 4x + 3x + 4

⇒ 3x2 + 7x + 4 – 2x2 – 11x – 15 = 0

⇒ x2 – 4x – 11 = 0

Here a = 1, b = - 4, c = - 11

D = b2 – 4ac

= (- 4)2 – 4 × 1 × (-11)

= 16 + 44

= 60 (ii) 2/(x + 2) – 1/(x + 1) = 4/(x + 4) – 3/(x + 3)

2/(x + 2) – 1/(x + 1) = 4/(x + 4) – 3/(x + 3)

(2x + 2 – x – 2)/{(x + 2)(x + 1) = (4x + 12 – 3x – 12)/{(x + 4)(x + 3)}

⇒ x/{(x + 2)(x + 1)} = x/{(x + 4)(x + 3)}

⇒ 1/{(x + 2)(x + 1)} = 1/{(x + 4)(x + 3)}[Dividing by x if x ≠ 0]

⇒ 1//(x2 + 3x + 2) = 1/(x2 + 7x + 12)

⇒ x2 + 7x + 12 – x2 – 3x – 2 = 0

⇒ 4x + 10

⇒ 2x + 5 = 0

⇒ 2x = - 5

⇒ x = - 5/2

If x = 0, then

0/{(x + 2)(x + 1) = 0/{(x + 4)(x + 3)} which is correct

Hence, x = 0, - 5/2

7. (i) (3x – 4)/7 + 7/(3x – 4), x 4/3

(ii) 4/x – 3 = 5/(2x + 3), x 0, - (3/2)

(i) (3x – 4)/7 + 7/(3x – 4) = 5/2, x ≠ 4/3

Let (3x – 4)/7 = y, then

y + 1/y = 5/2

⇒ 2y2 + 2 = 5y

⇒ 2y2 – 5y + 2 = 0

⇒ 2y2 – y – 4y + 2 = 0

⇒ y(2y – 1) – 2(2y – 1) = 0

⇒ (2y - 1)(y – 2) = 0

Either 2y – 1 =0, then

2y = 1

⇒ y = 1/2

Or y – 2 = 0, then y = 2

When y = 1/2, then

(3x – 4)/7 = 1/2

⇒ 6x – 8 = 7

⇒ 6x = 7 + 8

⇒ 6x = 15

⇒ x = 15/6

= 5/2

y = 2, then

(3x – 4)/7 = 2/1

⇒ 3x – 4 = 14

⇒ 3x = 14 + 4 = 18

⇒ x = 18/3 = 6

∴ x = 6, 5/2

(ii) 4/x – 3 = 5/(2x + 3), x ≠ 0, - (3/2)

⇒ (4 – 3x)/x = 5/(2x + 3)

⇒ (4 – 3x)(2x + 3) = 5x

⇒ 8x + 12 – 6x2 – 9x – 5x = 0

⇒ - 6x2 – 6x + 12 = 0

⇒ x2 + x – 2 = 0

⇒ x2 + 2x – x – 2 = 0

⇒ x(x + 2) – 1(x + 2) = 0

⇒ (x + 2)(x – 1) = 0

Either x + 2 = 0, then x = - 2

Or x – 1 = 0, then x = 1

∴ x = 1, - 2

8. (i) x2 + (4 – 3a)x – 12a = 0

(ii) 10ax2 – 6x + 15ax – 9 0, a 0

(i) x2 + (4 – 3a)x – 12a = 0

Here a = 1, b = 4 - 3a, c = - 12a

∴ d = b2 – 4ac

= (4 – 3a)2 – 4 × 1 × (-12a)

= 16 – 24a + 9a2 + 48a

= 16 + 24a + 9a2

= (4 + 3a)2 = (3a – 4 ± 3a + 4)/2

∴ x1 = (3a – 4 + 3a + 4)/2

= 6a/2

= 3a

And x2 = (3a – 4 – 3a – 4)/2

= (- 8)/2

= - 4

∴ Roots are 3a, - 4 s

(ii) 10ax2 – (6 – 15a) x – 9 = 0

Here a = 10a, b – (6 – 15a), c = - 9

d = b2 – 4ac

∴ [- (6 – 15a)]2 – 4 × 10a(-9)

= 36 – 180a + 225a2 + 360a

= 36 + 180a + 225a2

= 3/5a

x2 = (6 – 15a – 6 – 15a)/(20a)

= (-30a)/(20a)

= -3/2

Hence, x = 3/5a, - 3/2

9. Solve for x using the quadratic formula. Write your answer correct to two significant figures: (x – 1)2 – 3x + 4 = 0.

(x – 1)2 – 3x + 4 = 0

x2 + 1 – 2x - 3x + 4 = 0

x2 – 5x + 5 = 0

Here a = 1, b = - 5 and c = 5 = (5 + 2.2236)/2 or (5 – 2.236)/2

= (7.236)/2 or (2.764)/2

= 3.618 or 1.382

∴ x = 3.6, 1.4

10. Discuss the nature of the roots of the following equations:

(i) 3x2 – 7x + 8 = 0

(ii) x2 – 1/2.x – 4 = 0

(i) 3x2 – 7x + 8 = 0

Here, a = 3, b = - 7, c = 8

∴ D = b2 – 4ac

= (-7)2 – 4 × 3 × 8

= 49 – 96

= - 47

∵ D < 0

∴ Roots are not real

(ii) x2 – 1/2.x – 4 = 0

Here a = 1, b = 1/2, c = - 4

∴ D = b2 – 4ac

= (1/2)2

= 4 × 1 × (-4)

= 1/4 + 16

= 65/4

∵ D > 0

∴ Roots are real and distinct

(iii) 5x2 - 65x + 9 = 0

Here a = 5, b = - 6√5, c = 9

∴ D = b2 – 4ac

= (- 6√5)2 - 4 × 5 × 9

= 180 – 180

= 0

∴ D = 0

∴ Roots are real and equal.

(iv) √3x2 – 2x - √3 = 0

Here a = √3, b = - 2, c = - √3

∴ D = b2 – 4ac

= (- 2)2 – 4 × √3 × (-√3)

= 4 + 12

= 16

∵ D > 0

∴ Roots are real and distinct.

11. Find the values of k so that the quadratic equation (4 – k)x2 + 2(k + 2)x + (8k + 1) = 0 has equal roots.

(4 – k)x2 + 2(k + 2)x + (8k + 1) = 0

Here a = (4 – k), b = 2(k + 2), c = 8k + 1

∴ D = b2 – 4ac

= [2(k + 2)]2 – 4 × (4 – k)(8k + 1) = 0

= 4(k + 2)2 – 4(32k + 4 – 8k2 – k)

= 4(k2 + 4k + 4) – 4(32k + 4 – 8k2 – k)

= 4k2 + 16k + 16 – 128k – 16 + 32k2 + 4k

= 36k2 – 108k

= 36k(k – 3)

∵ Roots are real.

∴ D = 0

⇒ 36k(k – 3) = 0

⇒ k(k – 3) = 0

Either k = 0

Or k – 3 = 0, then k = 3

k = 0, 3

12. Find the values of m so that the quadratic equation 3x2 – 5x – 2m = 0 has two distinct real roots.

3x2 – 5x – 2m = 0

Here a = 3, b = - 5, c = - 2m

∴ D = b2 – 4ac

= (-5)2 – 4 × 3 × (-2m)

= 25 + 24m

24m > - 25

m > - (25/24)

13. Find the value(s) of k which each of the following quadratic equation has equal roots:

(i) 3kx2 = 4(kx – 1)

(ii) (k + 4)x2 + (k + 1)x + 1 = 0

Also, find the roots for that value (s) of k in each case.

(i) 3x2 = 4(kx – 1) = 0

⇒ 3kx2 = 4kx – 4

⇒ 3kx2 – 4kx + 4 = 0

Here a = 3k, b = - 4k, c = 4

∴ D = b2 – 4ac

= (-4k)2 – 4 × 3k × 4

= 16k2 – 48k

∴ Roots are equal

∴ D = 0

⇒ 16k2 – 48k = 0

⇒ k2 – 3k = 0

⇒ k(k – 3) = 0

Either k = 0 or k – 3 = 0

or k – 3 = 0

⇒ then k = 3

∴ x = (- b ± )/2a

= - b/(2a) (∵ D = 0)

4k/(2 × 3k)

= (4 × 3)/(2 × 3 × 3)

= 112/18

= 2/3

∴ x = 2/3, 2/3

(ii) (k + 4)x2 + (k + 1)x + 1 = 0

Here a = k + 4, b = k + 1, c = 1

∴ D = b2 – 4ac

= (k + 1)2 – 4 × (k + 4) × 1

= k2 + 2k + 1 – 4k – 16

= k2 – 2k – 15

∵ Root are equal.

∴ k2 – 2k – 15 = 0

⇒ k2 – 5k + 3k – 15 = 0

⇒ k(k – 5) + 3(k – 5) = 0

Either k – 5 = 0, then k = 5

or k + 3 = 0, then k = - 3

(a) When k = 5, then

x = (- b ± )/2a

= - b/2a

= - k – 1)/{2(k + 4)}

= (- 5 – 1)/{2(5 + 4)}

= (-6)/18

= (-1)/3

∴ x = - 1/3, -1/3

(b) When k = - 3, then

x = (- b ± )/2a

= -b/2a

= (- k – 1)/{2(k + 4)}

= {(-3) – 1}/{2(-3 + 4)}

= 2/(2 × 1)

= 1

∴ x = 1, 1

14. Find two natural numbers which differ by 3 and whose squares have the sum 117.

Let first natural number = x

Then second natural number = x + 3

According to the condition:

x2 + (x + 3)2 = 117

⇒ x2 + x2 + 6x + 9 = 117

⇒ 2x2 + 6x + 9 – 117 = 0

⇒ 2x2 + 6x – 108 = 0

⇒ x2 + 3x – 54 = 0 (Dividing by 2)

⇒ x2 + 9x – 6x – 54 = 0

⇒ x(x + 9) – 6(x + 9) = 0

⇒ (x + 9)(x – 6) = 0

Either x + 9 = 0, then x = - 9, but it is not a natural number.

Or x – 6 = 0, then x = 6

∴ First natural number = 6

And second number = 6 + 3

= 9

15. Divide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by 164.

Let larger part = x

Then smaller part = 16 – x

(∵ sum = 16)

According to the condition

2x2 – (16 – x)2 = 164

⇒ 2x2 – (256 – 32x + x2) = 164

⇒ 2x2 – 256 + 32x – x2 = 164

⇒ x2 + 32x – 256 – 164 = 0

⇒ x2 + 32x – 420 = 0

⇒ x2 + 42x – 10x – 420 = 0

⇒ x(x + 42) – 10(x + 42) = 0

⇒ (x + 42)(x – 10) = 0

Either x + 42 = 0, then x = - 42, but it is not possible.

Or x – 10 = 0, then x = 10

∴ Larger part = 10

And smaller part = 16 – 10 = 6

16. Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.

Ratio in two numbers = 3 : 4

Let the numbers be 3x and 4x

According to the condition,

(4x)2 – (3x)2 = 175

⇒ 16x2 – 9x2 = 175

⇒ 7x2 = 175

⇒ x2 = 175/7

= 25

= (± 5)2

∴ x = 5

∴ Natural numbers are 3x, 4x

= 3 × 5, 4 × 5

= 15, 20

17. Two squares have sides A cm and (x + 4) cm. The sum of their areas is 656 sq. cm. Express this as an algebraic equation and solve it to find the sides of the squares.

Side of first square = x cm

And side of second square = (x + 4) cm

Now according to the condition,

(x)2 + (x + 4)2 = 656

⇒ x2 – x2 + 8x + 16 = 656

⇒ 2x2 + 8x + 16 – 656 = 0

⇒ 2x2 – 8x – 640 = 0

⇒ x2 + 20x – 16x – 320 = 0

⇒ x(x + 20) – 16(x + 20) = 0

⇒ (x + 20)(x – 16) = 0

Either x + 20 = 0, then x = - 20, but it is not possible as it is in negative.

Or x – 16 = 0 then x = 16

Side of first square = 16 cm

And side of second square = 16 + 4 – 4

= 20 cm

18. The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.

Then length = (x + 12) m

Area = l × b

= x(x + 12) m2

And perimeter = 2(l + b)

= 2(x + 12 + x)

= 2(2x + 12) m

According to the condition,

x(x + 12) = 4 × 2(2x + 12)

⇒ x2 + 12x = 16x + 96

⇒ x2 + 12x – 16x – 96 = 0

⇒ x2 – 4x – 96 = 0

⇒ x2 – 12x + 8x – 96 = 0

⇒ x(x – 12) + 8(x – 12) = 0

⇒ (x – 12)(x + 8) = 0

Either x – 12 = 0, then x = 12

Or x + 8 = 0, then x = - 8, but it is not possible as it is in negative.

And length = 12 + 12

= 24 m

19. A farmer wishes to grow a 100m2 rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.

Area of rectangular = 100 cm2

Length of barbed wire = 30 m

Let the length of the side opposite to wall = x

And length of the side opposite to wall = x

And length of other each side = (30 – x)/2

According to the condition, x(30 – x)/2 = 100

⇒ x(30 – x)= 200

⇒ 30x – x2 = 200

⇒ x2 – 30x + 200 = 0

⇒ x2 – 20x – 10x + 200 = 0

⇒ x(x – 20) – 10(x – 20) = 0

⇒ (x – 20)(x – 10) = 0

Either x – 20 = 0, then x = 20

Or x – 10 = 0, then x = 10

(i) If x = 20, then side opposite to the wall = 20 m

and other side = (30 – 20)/2

= 10/2

= 5 m

(ii) If x = 10, then side opposite to wall = 10 m

and other side = (30 – 10)/2

= 20/2

= 10 m

∴ Sides are 20 m, 5m or 10m, 10 m

20. The hypotenuse of a right-angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.

Let the length of shortest side = x m

Length of hypotenuse = 2x – 1

And third side = x + 1

Now according to the condition,

(2x – 1)2 = (x)2 + (x + 1)2 (By Pythagoras theorem)

⇒ 4x2 – 4x + 1 = x2 + x2 + 2x + 1

⇒ 4x2 – 4x + 1 – 2x2 – 2x – 1 = 0

⇒ 2x2 – 6x = 0

⇒ x2 – 3x = 0 (Dividing by 2)

⇒ x(x – 3) = 0

Either x = 0, but it is not possible

Or x – 3 = 0, then x = 3

∴ Shortest side = 3 m

Hypotenuse = 2 × 3 – 1

= 6 – 1

= 5

Third side = x + 1

= 3 + 1

= 4

Hence sides are 3, 4, 5 (in m)

21 A wire; 112 cm long is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.

Perimeter of a right angled triangle = 112 cm

Hypotenuse = 50 cm

∴ Sum of other two sides = 112 – 50

= 62 cm

Let the length of first side = x

And length of other side = 62 – x

According to the condition,

(x)2 + (62 – x)2 = (50)2 [By Pythagoras theorem]

⇒ x2 + 3844 – 124x + x2 = 2500

⇒ 2x2 – 124x + 1344 – 2500 = 0

⇒ 2x2 – 124x + 1344 = 0

⇒ x2 – 62x + 672 = 0 (Dividing by 2)

⇒ x2 – 48x – 14x + 672 = 0

⇒ x(x – 48) – 14(x – 48) = 0

⇒ (x – 48)(x – 14) = 0

Either x – 48 = 0, then x = 49

or x – 14 = 0, then x = 14

(i) If x = 48, then one side = 48 cm

And other side = 62 – 48

= 14 cm

(ii) If x = 14, then one side = 14 cm

And other side = 62 – 14

= 48

Hence sides are 14 cm, 48 cm

22. Car A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.

(i) Write down the number of litres used by car A and car B in covering a distance of 400 km.

(ii) If car A uses 4 litres of petrol more than car B in covering 400 km. Write down an equation, in A and solve it to determine the number of litres of petrol used by car B for the journey.

Distance travelled by car A in one litre = x km

And distance travelled by car B in one litre = (x + 5) km

(i) Consumption of car A in covering 400 km = 400/x litres and consumption of car B

= 400/(x + 5) litres.

(ii) According to the condition, 400/x – 400/(x + 5) = 4

= 400(x + 5 – x)/{x(x + 5)}

= 4

⇒ (400 × 5)/(x2 + 5x) = 4

⇒ 2000 = 4x2 + 20x

⇒ 4x2 + 20x – 2000 = 0

⇒ x2 + 5x – 500 = 0 (Dividing by 4)

⇒ x2 + 25x – 20x – 500 = 0

⇒ x(x + 25) – 20(x + 25) = 0

⇒ (x + 25)(x – 20) = 0

Either x + 25 = 0, then x = - 25, but it is not possible as it is in negative.

or x – 20 = 0, then x = 20

∴ Petrol used by car B = 20 – 4

= 16 litres.

23. The speed of a boat in still water is 11 km/hr. It can go 12 km up-stream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream.

Speed of a boat in still water = 11 km/hr

Let the speed of stream = x km/hr

Distance covered = 12 km

Time taken = 2 hours 45 minutes

= 2.3/4 = 11/4 hours

Now according to the condition

12/(11 – x) + 12/(11 + x) = 11/4

⇒ 12(11 + x + 11 – x)/{(11 – x)(11 + x) = 11/4

⇒ (12 × 22)/(121 – x2) = 11/4

⇒ 1331 – 11x2 = 4 × 12 × 22 = 1056

⇒ 1331 – 11x2 = 1056

⇒ 1331 – 1056 – 11x2 = 0

⇒ - 11x2 + 275 = 0

⇒ x2 – 25 0 (Dividing by – 11)

⇒ (x + 5)(x – 5) = 0

Either x + 5 = 0, then x = - 5, but it is not possible as it is in negative.

or x – 5 = 0, then x = 5

Hence speed of stream = 5 km/hr.

24. By selling an article for Rs 21, a trader loses as much per cent as the cost price of the article. Find the cost price.

S.P. of an article = Rs 21

Let cost price = Rs x

Then loss = x%

Then loss = x%

Then loss = x %

∴ S.P. = C.P. (100 – Loss %)/10

21 = x(100 – x)/100

2100 = 100x – x2

⇒ x2 – 100x + 2100 = 0

⇒ x2 – 30x – 7x + 2100 = 0

⇒ x(x – 30) – 70(x – 30) = 0

⇒ (x – 30) (x – 70) = 0

Either x – 30 = 0, then x = 30

Or x – 70 = 0, then x = 70

∴ Cost price = Rs 30 or Rs 70

25. A man spent Rs 2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by Rupee 1. The man finally paid Rs 2730and received 1 more plants. Find x.

Amount spent = Rs 2800

Price of each plant = Rs x

Reduced price = Rs (x – 1)

No. of plants in first case = 2800/x

No. of plants received in second case = 2800/x + 10

Amount paid = Rs 2730

According to the condition,

(2800/x + 10)(x – 1) = 2730

⇒ {(2800 + 10x)(x – 1)}/x = 2730

⇒ (2800 + 10x)(x – 1) = 2730x

⇒ 2800x – 2800 + 10x2 – 10x = 2730x

⇒ 10x2 + 2800x – 10x – 2730x – 2800 = 0

⇒ 10x2 + 60x – 2800 = 0

⇒ x2 + 6x – 280 = 0 (Dividing by 10)

⇒ x2 + 20x – 14x – 280 = 0

⇒ x(x + 20) – 14(x + 20) = 0

⇒ (x + 20)(x – 14) = 0

Either x + 20 = 0, then x = - 20, but it is not possible as it is in negative.

Or x – 14 = 0, then x = 14

26. Forty years hence, Mr. Pratap’s age will the square of what it was 32 years years ago. Find his present age.

Let Pratap’s present age = x years

40 years hence his age = x + 40

And 32 years ago his age = x – 32

According to the condition

x + 40 = (x – 32)2

⇒ x + 40 = x2 – 64x + 1024

⇒ x2 – 64x + 1024 – x – 40 = 0

⇒ x2 – 65x + 984 = 0

⇒ x2 – 24x – 41x + 984 = 0

⇒ x(x - 24) – 41(x – 24) = 0

⇒ (x – 24)(x – 41) = 0

Either x – 24 = 0, then x = 24 but it is not possible as it is less than 32

Or x – 41 = 0, then x = 41

Hence present age = 41 years

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