# RS Aggarwal Solutions Chapter 10 Quadratic Equation Exercise 10D Class 10 Maths

 Chapter Name RS Aggarwal Chapter 10 Quadratic Equation Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 10AExercise 10BExercise 10CExercise 10EExercise 10F Related Study NCERT Solutions for Class 10 Maths

### Exercise 10D Solutions

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

(i) 2x2 – 8x + 5 = 0

(ii) 3x2 - 2√6x + 2 = 0

(iii) 5x2 – 4x + 1 = 0

(iv) 5x(x – 2) + 6 = 0

(iv) 5x(x – 2) + 6 = 0

(v) 12x2 + 4√15 + 5 = 0

(vi) x2 – x + 2 = 0

Solution

(i) The given equation is 2x2 – 8x + 5 = 0

This is of the form ax2 + bx + c = 0, where a = 2, b = - 8 and c = 5.

∴ Discriminant, D = b2 – 4ac = (-8)2 – 4 × 2 × 5 = 64 – 40

= 24 > 0

Hence, the given equation has real and unequal roots.

(ii) The given equation is 3x2 - 2√6x + 2 = 0

This is of the form ax2 + bx + c = 0, where a = 3, b = -2√6 and c = 2

∴ Discriminant, D = b2 – 4ac

= (-2√6)2 – 4 × 3 × 2

= 24 – 24

= 0

Hence, the given equation has real and equal roots.

(iii) The given equation is 5x2 – 4x + 51 = 0.

This is of the form ax2 + bx + c = 0, where a = 5, b = - 4 and c = 1.

∴ Discriminant,

D = b2 – 4ac

= (-4)2 – 4 × 5 × 1’

= 16 – 20

= - 4 < 0

Hence, the given equation has no real roots.

(iv) The given equation is

5x(x – 2) + 6 = 0

⇒ 5x2 – 10x + 6 = 0

This is of the form ax2 + bx + c = 0, where a = 5, b = - 10 and c = 6.

∴ Discriminant. D = b2 – 4ac

= (-10)2 – 4 × 5 × 6

= 100 – 120

= -20 < 0

Hence, the given equation has no real roots.

(v)

(vi) The given equation is x2 – 2x + 2 = 0

This is of the form ax2 + bx + c = 0, where a = 1, b = - 1 and c = 2.

∴ Discriminant, D = b2 – 4ac = {(-1)2 – 4 × 1 × 2}

= 1 – 8

= -7 < 0

Hence, the given equation has no real roots.

2. If a and b are distinct real numbers, show that the quadratic equations 2(a2 + b2)x2 + 2(a + b)x + 1 = 0. has no real roots.

Solution

The given equation is 2(a2 + b2)x2 + 2(a + b)x + 1 = 0.

∴ D = [2(a + b)]2 – 4 × 2(a2 + b2) × 1

= 4(a2 + 2ab + b2) – 8(a2 + b2)

= 4a2 + 8ab + 4b2 – 8a2 – 8b2

= -4a2 + 8ab – 4b2

= -4(a2 – 2ab + b2)

= -4(a – b)2 < 0

Hence, the given equation has no real roots.

3. Show that the roots of the equation x2 + px – q2 = 0 are real for all real values of p and q.

Solution

x2 + px – q2 = 0

Here,

a = 1, b = p and c = - q2

Discriminant D is given by:

D = (b2 – 4ac)

= p2 - 4 × 1 × (-q)2

= (p2 + 4q2) > 0

D > 0 for all real values of p and q.

Thus, the roots of the equation are real.

4. For what values of k are the roots of the quadratic equation 3x2 + 2kx + 27 = 0 real and equal ?

Solution

3x2 + 2kx + 27 = 0

Here,

a = 3, b = 2k and c = 27

It is given that the roots of the equation are real and equal; therefore, we have:

D = 0

⇒ (2k)2 – 4 × 3 × 27 = 0

⇒ 4k2 – 324 = 0

⇒ 4k2 = 324

⇒ k2 = 81

⇒ k2 = ± 9

∴ k = 9 or k = -9

5. For what value of k are the roots of the quadratic equation kx(x - 2√5) + 10 = 0 real and equal.

Solution

The given equation is

kx(x - 2√5) + 10 = 0

⇒ kx2 - 2√5kx + 10 = 0

This is of the form ax2 + bx + c = 0, where a = k, b = -2√5k and c = 10.

∴ D = b2 – 4ac

= (-2√5k)2 – 4 × k × 10

= 20k2 – 40k

The given equation will have real and equal roots if D = 0.

∴ 20k2 – 40k = 0

⇒ 20k(k – 2) = 0

⇒ k = 0 or k – 2 = 0

⇒ k = 0 or k = 2

But, for k = 0, we get 10 = 0, which is not true

Hence, 2 is the required value of k.

6. For what values of p are the roots of the equation 4x2 + px + 3 = 0. real and equal?

Solution

The given equation is 4x2 + px + 3 = 0

This is of the form ax2 + bx + c = 0, where a = 4, b = p and c = 3.

∴ D = b2 – 4ac

= p2 – 4 × 4 × 3

= p2 – 48

The given equation will have real and equal roots if D = 0.

∴ p – 48 = 0

⇒ p2 = 48

Hence, 4√3 and -4√3 are the required values of p.

7. Find the nonzero value of k for which the roots of the quadratic equation 9x2 – 3kx + k = 0.

Solution

The given equation is 9x2 – 3kx + k = 0.

This is of the form ax2 + bx + c = 0, where a = 9, b = - 3k and c = k.

∴ D = b2 – 4ac

= (-3k)2 – 4 × 9 × k

= 9k2 – 36k

The given equation will have real and equal roots is D = 0.

∴ 9k2 – 36k = 0

⇒ 9k(k – 4) = 0

⇒ k = 0 or k – 4 = 0

⇒ k = 0 or k = 4

But, k ≠ 0 (Given)

Hence, the required values of k is 4.

8Find the values of k for which the quadratic equation (3k + 1)x2 + 2(k + 1)x + 1 = 0. has real and equal roots.

Solution

The given equation is (3k + 1)x2 + 2(k + 1)x + 1 = 0.

This is of the form ax2 + bx + c = 0, where a = 3k + 1, b = 2(k + 1) and c = 1.

∴ D = b2 – 4ac

= [2(k + 1)]2 – 4 × (3k + 1) × 1

= 4(k2 + 2k + 1) – 4(3k +1)

= 4k2 + 8k + 4 – 12k – 4

= 4k2 – 4k

The given equation will have real and equal roots if D = 0.

∴ 4k2 – 4k = 0

⇒ 4k(k – 1) = 0

⇒ k = 0 or k – 1 = 0

⇒ k = 0 or k = 1

Hence, 0 and 1 are the required values of k.

9. Find the value of p for which the quadratic equation (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0. has real and equal roots.

Solution

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

This is of the form ax2 + bx + c = 0, where a = 2p + 1, b = -(7p + 2) and c = 7p – 3

∴ D = b2 – 4ac

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

= (49p2 + 28p + 4) - 4(14p2 + p – 3)

= 49p2 + 28p + 4 – 56p2 – 4p + 12

= -7p2 + 24p + 16

The given equation will have real and equal roots if D = 0.

∴ -7p2 + 24p + 16 = 0

⇒ 7p2 – 24p – 16 = 0

⇒ 7p2 – 28p + 4p – 16 = 0

⇒ 7p(p – 4) + 4(p – 4) = 0

⇒ (p – 4)(7p + 4) = 0

⇒ p – 4 = 0 or 7p + 4 = 0

⇒ p = 4 or p = -(4/7)

Hence, 4 and –(4/7) are the required values of p.

10. Find the values of p for which the quadratic equation (p + 1)x2 – 6(p + 1)x + 3(x + 9) = 0., p ≠ - 1has equal roots. hence, find the roots of the equation.

Solution

The given equation is (p + 1)x2 – 6(p + 1)x + 3(p + 9) = 0

This is of the form ax2 + bx + c = 0, where a = p + 1, b = -6(p + 1) and c = 3(p + 9).

∴ D = b2 – 4ac

= [-6(p+ 1)]2 – 4 × (p + 1) × 3(p + 9)

= 12(p + 1)[3(p + 1) – (p + 9)]

= 12(p + 1)(2p – 6)

The given equation will have real and equal roots if D = 0.

∴ 12(p + 1)(2p – 6) = 0

⇒ p + 1 = 0 or 2p – 6 = 0

⇒ p = -1 or p = 3

But, p ≠ -1 (Given)

Thus, the value of p is 3

Putting p = 3, the given equation becomes 4x2 – 24x + 36 = 0

4x2 – 24x + 36 = 0

⇒ 4(x2 – 6x + 9) = 0

⇒ (x – 3)2 = 0

⇒ x – 3 = 0

⇒ x = 3

Hence, 3 is the repeated root of this equation.

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

Solution

It is given that -5 is a root of the quadratic equation 2x2 + px - 15 = 0.

∴ 2(-5)2 + p × (-5) – 15 = 0

⇒ - 5p + 35 = 0

⇒ p = 7

The roots of the equation px2 + px + k = 0 = 0 are equal.

∴ D = 0

⇒ p2 – 4pk = 0

⇒ (7)2 – 4 × 7 × k = 0

⇒ 49 – 28k = 0

⇒ k = 49/28 = 7/4

Thus, the value of k is 7/4.

12. If 3 is a root of the quadratic equation x2 – x + k = 0., find the value of p so that the roots of the equation x2 + 2kx + (k2 + 2k + p) = 0 are equal.

Solution

It is given that 3 is a root of the quadratic equation x2 – x + k = 0

∴ (3)2 – 3 + k = 0

⇒ k + 6 = 0

⇒ k = - 6

The roots of the equation x2 + 2kx + (k2 + 2k + p) = 0 are equal.

∴ D = 0

⇒ (2k)2 – 4 × 1 × (k2 + 2k + p) = 0

⇒ 4k2 – 4k2 – 8k – 4p = 0

⇒ - 8k – 4p = 0

⇒ p = 8k/-4 = - 2k

⇒ p = 8k/-4 = -2k

⇒ p = -2 × (-6) = 12

Hence, the value of p is 12.

13. If -4 is a root of the equation x2 + 2x + 4p = 0. find the value of k for the which the quadratic equation x2 + px(1 + 3k) + 7(3 + 2k) = 0 has real roots.

Solution

It is given that -4 is a root of the quadratic equation x2 + 2x + 4op = 0

∴ (-4)2 + 2 × (-4) + 4p = 0

⇒ 16 – 8 + 4p = 0

⇒ 4p + 8 = 0

⇒ p = - 2

The equation x2 + px(1 + 3k) + 7(3 + 2k) = 0 has real roots.

∴ D = 0

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

⇒ [-2(1 + 3k)]2 – 28(3 + 2k) = 0

⇒ 4(1 + 6k + 9k2) – 28(3 + 2k) = 0

⇒ 4(1 + 6k + 9k2 – 21 – 14k) = 0

⇒ 9k2 – 8k – 20 = 0

⇒ 9k2 – 18k + 10k – 20 = 0

⇒ 9k(k – 2) + 10(k – 2) = 0

⇒ (k – 2)(9k + 10) = 0

⇒ k – 2 = 0 or 9k + 10 = 0

⇒ k = 2 or k = -(10/9)

Hence, the required value of k is 2 or –(10/9).

14. If the quadratic equation (1 + 4m2)x2/ + 2mcx + (c2 – a2) = 0 has equal roots, prove that c2 = a2 (1 + m2).

Solution

Given:

(1 + m2)x2 + 2mcx + (c2 – a2) = 0

Here,

a = (1 + m2), b = 2mc and c = (c2 – a2)

It is given that the roots of the equation are equal; therefore, we have:

D = 0

⇒ (b2 – 4ac) = 0

⇒ (2mc)2 – 4 × (1 + m2) × (c2 – a2) = 0

⇒ 4m2c2 – 4(c2 – a2 + m2c2 - m2a2) = 0

⇒ 4m2c2 – 4c2 + 4a2 – 4m2c2 + 4m2a2 = 0

⇒ - 4c2 + 4a2 + 4m2a2 =0

⇒ a2 + m2a2 = c2

⇒ a2(1 + m2) = c2

⇒ c2 = a2(1 + m2)

Hence proved.

15. If the roots of the quadratic equation (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0 are real and equal, show that either a = 0 or (a3 + b3 + c3 = 3abc)

Solution

Given:

(c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0

Here,

a = (c2 – ab), b = - 2(a2 – bc), c = (b2 – ac)

It is given that the roots of the equation are real and equal; therefore, we have:

D = 0

⇒ (b2 – 4ac) = 0

⇒ {-2(a2 – bc)}2 – 4 × (c2 – ab) × (b2 – ac) = 0

⇒ 4(a4 – 2a2bc + b2c2) – 4(b2c2 – ac3 – ab3 + a2bc) = 0

⇒ a4 – 2a2bc + b2c2 – b2c2 + ac3 + ab3 – a2bc = 0

⇒ a4 – 3a2bc + ac3 + ab3 = 0

⇒ a(a3 – 3abc + c3 + b3) = 0

Now,

a = 0 or a3 – 3abc + c3 + b3 = 0

a = 0 or a3 + b3 + c3 = 3abc

16. Find the value of p for which the quadratic equation 2x2 + px + 8 = 0 has real roots.

Solution

Given:

2x2 + px + 8 = 0

Here,

a = 2, b = p and c = 8

Discriminant D is given by

D = (b2 – 4ac)

= p2 – 4 × 2 × 8

= (p2 – 64)

If D ≥ 0, the roots of the equation will be real

⇒ (p2 – 64) ≥ 0

⇒ (p + 8)(p – 8) ≥ 0

⇒ p ≥ 8 and p ≤ - 8

Thus, the roots of the equation are real for p ≥ 8 and p ≤ -8.

17. Find the value of a for which the equation (α – 12)x2 + 2(α – 12)x + 2 = 0 has equal roots.

Solution

(α – 12)x2 + 2(α – 12x + 2 = 0

Here,

a = (α – 12), b = 2(α – 12) and c = 2

It is given that the roots of the equation are equal; therefore, we have

D = 0

⇒ (b2 – 4ac) = 0

⇒ {2(α – 12)}2 – 4 × (α – 12) × 2 = 0

⇒ 4(α2 - 24α + 144) - 8α + 96 = 0

⇒ 4α2 - 96α + 576 - 8α + 96 = 0

⇒ 4α2 - 104α + 672 = 0

⇒ α2 - 26α + 168 = 0

⇒ α2 - 14α - 12α + 168 = 0

⇒ α(α – 14) – 12(α – 14) = 0

⇒ (α – 14)(α – 12) = 0

∴ α = 14 or α = 12

If the value of α is 12, the given equation becomes non-quadratic.

Therefore, the value of α will be 14 for the equation to have equal roots.

18. Find the value of k for which the roots of 9x2 + 8kx + 16 = 0 are real and equal

Solution

Given:

9x2 + 8kx + 16 = 0

Here,

a = 9, b = 8k and c = 16

It is given that the roots of the equation are real and equal; therefore, we have:

D = 0

⇒ (b2 – 4ac) = 0

⇒ (8k)2 – 4 × 9 × 16 = 0

⇒ 64k2 – 576 = 0

⇒ 64k2 = 576

⇒ k2 = 9

⇒ k = ± 3

∴ k = 3 or k = -3

19. Find the values of k for which the given quadratic equation has real and distinct roots:

(i) kx2 + 6x + 1 = 0.

(ii) x2 – kx + 9 = 0

(iii) 9x2 + 3kx + 4 = 0

(iv) 5x2 – kx + 1 = 0

Solution

(i) The given equation is kx2 + 6x + 1 = 0.

∴ D = 62 – 4 × k × 1 = 36 – 4k

The given equation has real and distinct roots if D > 0.

∴ 36 – 4k > 0

⇒ 4k < 36

⇒ k < 9

(ii) The given equation is x2 – kx + 9 = 0

∴ D = (-k)2 – 4 × 1 × 9

= k2 – 36

The given equation has real and distinct roots if D > 0

∴ k2 – 36 > 0

⇒ (k – 6)(k + 6) > 0

⇒ k < -6 or k > 6

(iii) the given equation is 9x2 + 3kx + 4 = 0

∴ D = (3k)2 – 4 × 9 × 4

= 9k2 – 144

The given equation has real and distinct roots if D > 0.

∴ 9k2 – 144 > 0

⇒ 9(k2 – 16) > 0

⇒ (k – 4)(k + 4) > 0

⇒ k < - 4 or k > 4

(iv) The given equation is 5x2 – kx + 1 = 0.

∴ D = (-k)2 – 4 × 5 × 1

= k2 – 20

The given equation has real and distinct roots if D > 0.

∴ k2 – 20 > 0

⇒ k2 - (2√5)2 > 0

⇒ (k - 2√5)(k + 2√5) > 0

⇒ k = -2√5 or k > 2√5

20. If a and b are real and a ≠ b then show that the roots of the equation

(a – b)x2 + 5(a + b)x – 2(a – b) = 0 are equal and unequal.

Solution

The given equation is (a – b)x2 + 5(a + b)x – 2(a – b) = 0.

∴ D = [5(a + b)]2 – 4 × (a – b) × [-2(a – b)]

= 25(a + b)2 + 8(a – b)2

Since a and b are real and a ≠ b, so (a – b)2 > 0 and (a + b)2 > 0

∴ 8(a – b)2 > 0 ...(1) (Product of two positive numbers is always positive)

Also, 25(a + b)2 > 0 ...(2) (Product of two positive numbers is always positive)

Adding (1) and (2), we get

25(a + b)2 + 8(a – b)2 > 0 (Sum of two positive numbers is always positive)

⇒ D > 0

Hence, the roots of the given equation are real and unequal.

21. If the roots of the equation (a2 + b2)x2 – 2(ac + bd)x = (c2 + d2) = 0 are equal, prove that a/b = c/d.

Solution

It is given that the roots of the equation (a2 + b2)x2 - 2(ac + bd)x + (c2 + d2) = 0 are equal.

∴ D = 0

⇒ [-2(ac + bd)]2 – 4(a2 + b2)(c2 + d2) = 0

⇒ 4(a2c2 + b2d2 + 2abcd) – 4(a2c2 + a2d2 + b2c2 + b2d2) = 0

⇒ 4(a2c2 + b2d2 + 2abcd – a2c2 – a2d2 – b2c2 – b2d2) = 0

⇒ (-a2d2 + 2abcd – b2c2) = 0

⇒ -(a2d2 – 2abcd + b2c2) = 0

⇒ (ad – bc)2 = 0

⇒ ad – bc = 0

⇒ ad = bc

⇒ a/b = c/d

Hence proved.

22. If the roots of the equations ax2 + 2bx + c = 0 and bax2 - 2√acx + b = 0 are simultaneously real then prove that b2 = ac

Solution

It is given that the roots of the equation ax2 + 2bx + c = 0 are real.

∴ D1 = (2b)2 – 4 × a × c ≥ 0

⇒ 4(b2 – ac) ≥ 0

⇒ b2 – ac ≥ 0  ....(1)

⇒ 4(ac – b2) ≥ 0

⇒ - 4(b2 – ac) ≥ 0

⇒ b2 – ac ≥ 0 ...(2)

The roots of he given equations are simultaneously real if (1) and (2) holds true together.

this is possible if

b2 – ac = 0

⇒ b2 = ac