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

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

### Exercise 10B Solutions

1. x2 – 6x + 3 = 0

Solution

x2 – 6x + 3 = 0

⇒ x2 – 6x + 3 = 0

⇒ x2 – 6x = - 3

⇒ x2 – 2 × x × 3 + 32 = -3 + 32  (Adding 3on both sides)

⇒ (x – 3)2 = -3 + 9 = 6

⇒ x – 3 = ±√6  (Taking square root on the both sides)

⇒ x – 3 = √6 or x – 3 = -√6

⇒ x = 3 + √6 or x = 3 - √6

Hence, 3 + √6 and 3 - √6 are the roots of the given equation.

2. x2 – 4x + 1 = 0

Solution

x2 – 4x + 1 = 0

⇒ x2 – 4x = -1

⇒ x2 – 2 × x × 2 + 22 = -1 + 22 (Adding 22 on both sides)

⇒ (x – 2)2 = -1 + 4 = 3

⇒ x – 2 = ±√3 (Taking square root on both sides)

⇒ x – 2 = √3 or x – 2 = -√3

⇒ x = 2 + √3 or x = 2 - √3

Hence, 2 + √3 and 2 - √3 are the roots of the given equation.

3. x2 + 8x – 2 = 0

Solution

x+ 8x – 2 = 0

⇒ x2 + 8x = 2

⇒ x2 + 2 × x × 4 + 42 = 2 + 42 (Adding 42 on both sides)

⇒ (x + 4)2 = 2 + 16 = 18

⇒ x + 4 = ±√18 = ±3√2 (Taking square roots on the both sides)

⇒ x + 4 = 3√2 or x + 4 = -3√2

⇒ x = (-4 + 3√2) and x = (-4 - 3√2)

Hence, (- 4 + 3√2) and (-4 - 3√2) are the roots of the given equation.

4. 4x2 + 4√3x + 3 = 0

Solution

4x+ 4√3x + 3 = 0

⇒ 4x+ 4√3x = -3

⇒ (2x)2 + 2 × 2x × √3 + (√3)2

= - 3 + (√3)[Adding (√3)2 on both sides]

⇒ (2x + √3)= - 3 + 3 = 0

⇒ 2x + √3 = 0

⇒ x = -√3/2

Hence, -√3/2 is the repeated root of the given equation.

5. 2x2 + 5x – 3 = 0

Solution

2x+ 5x – 3 = 0

⇒ 4x2 + 10x – 6 = 0 (Multiplying both sides by 2)

⇒ 4x2 + 10x = 6

⇒ (2x2) + 2 × 2x × 5/2 + (5/2)2 = 6 + (5/2)2 [Adding (5/2)2 on both sides]

⇒ (2x + 5/2)2 = 6 + 25/4 = (24 + 25)/4 = 49/4 = (7/2)2

⇒ 2x + 5/2 = ± 7/2 (Taking square root on the both sides)

⇒ 2x + 5/2 = 7/2 or 2x + 5/2 = -(7/2)

⇒ 2x = 7/2 – 5/2 = 2/3 = 1 or 2x = -7/5 – 5/2

= - (12/2)

= -6

x = 1/2 or x = -3

Hence, 1/2 or x = -3 are the roots of the given equation.

6. 3x– x – 2 = 0

Solution

3x– x – 2 = 0

⇒ 9x2 – 3x – 6 = 0 (Multiplying both sides by 3)

⇒ 9x2 – 3x = 6

⇒ (3x)2 – 2 × 3x × 1/2 + (1/2)2

= 6 + (1/2)2 [Adding (1/2)2 on both sides]

⇒ (3x – 1/2)2 = 6 + 1/4 = 25/14 = (5/2)2

⇒ 3x – 1/2 = ± 5/2 (Taking square root on both sides)

⇒ 3x – 1/2 = 5/2 or 3x – 1/2 = -(5/2)

⇒ 3x = 5/2 + 1/2 = 6/2 = 3 or 3x = -5/2 + 1/2 = -(4/2) = -2

⇒ x = 1 or x = -(2/3)

Hence, 1 and –(2/3) are the roots of the given equation.

7. 8x2 - 14x – 15 = 0

Solution

8x2 – 14x – 15 = 0

⇒ 16x– 28x – 30 = 0 (Multiplying both sides by 2)

⇒ 16x2 – 28x = 30

⇒ (4x)2 – 2 × 4x × 7/2 + (7/2)2 = 30 + (7/2)2 [Adding (7/2)2 on both sides]

⇒ (4x – 7/2)= 30 + 49/4 = 169/4 = (13/2)2

⇒ 4x – 7/2 = ± 13/2 (Taking square root on both sides)

⇒ 4x – 7/2 = 13/2 or 4x – 7/2 = 13/2

⇒ 4x = 13/2 + 7/2 = 20/2 = 10 or 4x = -13/2 + 7/2 = -6/2 = -3

⇒ x = 5/2 or x = -3/4

Hence, 5/2 and -3/4 are the roots of the given equation.

8. 7x2 + 3x – 4 = 0

Solution

7x2 + 3x – 4 = 0

⇒ 49x2 + 21x – 28 = 0 (Multiplying both sides by 7)

⇒ 49x2 + 21x = 28

⇒ (7x)2 + 2 × 7x × 3/2 + (3/2)2 = 28 + (3/2)2 [Adding (3/2)2 on both sides]

⇒ (7x + 3/2)2 = 28 + 9/4 = 121/4 = (11/2)2

⇒ 7x + 3/2 = ± 11/2 (Taking square root on both sides)

⇒ 7x + 3/2 = 11/2 or 7x + 3/2 = -11/2

⇒ 7x = 11/2 – 3/2 = 8/2 = 4 or 7x = -11/2 – 3/2 = -14/2 = - 7

⇒ x = 4/7 or x = -1

Hence, 4/7 and -1 are the roots of the given equation.

9. 3x2 – 2x – 1 = 0

Solution

3x2 – 2x – 1 = 0

⇒ 9x2 – 6x – 3 = 0 (Multiplying both sides by 3)

⇒ 9x2 – 6x = 3

⇒ (3x)2 – 2 × 3x × 1 + 12 = 3 + 12 [Adding 12 on both sides]

⇒ (3x – 1)2 = 3 + 1 = 4 = (2)2

⇒ 3x – 1 = ± 2 (Taking square root on both sides)

⇒ 3x – 1 = 2 or 3x – 1 = -2

⇒ 3x = 3 or 3x = -1

⇒ x = 1 or x = -1/3

Hence, 1 and – 1/3 are the roots of the given equation.

10. 5x2 – 6x – 2 = 0

Solution

5x2 – 6x – 2 = 0

⇒ 25x2 – 30x -10 = 0 (Multiplying both sides by 5)

⇒ 25x2 – 30x = 10

⇒ (5x)2 – 2 × 5x × 3 + 32 = 10 + 32 (Adding 32 on both sides)

⇒ (5x – 3)2 = 10 + 9 – 19

11. 2/x2 – 5/x + 2 = 0

Solution

2/x2 – 5/x + 2 = 0

⇒ (2 – 5x + 2x2)/x2

⇒ 2x2 – 5x + 2 = 0

⇒ 4x2 – 10x + 4 = 0 (Multiplying both sides by 2)

⇒ 4x2 – 10x = -4

⇒ (2x)2 – 2 × 2x × 5/2 + (5/2)2 = - 4 + (5/2)2 [Adding (5/2)2 on both sides]

⇒ (2x – 5/2)2 = - 4 + 25/4 = 9/4 = (3/2)2

⇒ 2x – 5/2 = ± 3/2 (Taking square root on both sides)

⇒ 2x – 5/2 = 3/2 or 2x – 5/2 = -3/2

⇒ 2x = 3/2 + 5/2 = 8/2 = 4 or 2x = -3/2 + 5/2 = 2/2 = 1

⇒ x = 2 or x = 1/2

Hence, 2 and 1/2 are the roots of the given equation.

12. 4x2 + 4bx – (a2 – b2) = 0

Solution

4x2 + 4bx – (a2 – b2) = 0

⇒ 4x2 + 4bx = a2 – b2

⇒ (2x)2 + 2 × 2x × b + b2

= a2 – b2 + b2 (Adding b2 on both sides)

⇒ (2x + b)2 = a2

⇒ 2x + b = ±a (Taking square root on both sides)

⇒ 2x+ b = a or 2x + b = -a

⇒ 2x = a – b or 2x = - a – b

⇒ x = (a – b)/2 or x = -(a + b)/2

Hence, (a – b)/2 and –(a + b)/2 are the roots of the given equation.

13. x2 – (√2 + 1)x + √2 = 0

Solution

x2 – (√2 + 1)x + √2 = 0

⇒ x2 – (√2 + 1)x = -√2

⇒ x2 – 2 × x × (√2 + 1)/2 + ((√2 + 1)/2)2 = -√2 + ((√2 + 1)/2)2

[Adding ((√2 + 1)/2)2 on both sides]

⇒ [x – (√2 + 1)/2)2] = (-4√2 + 2 + 1 + 2√2)/4

= (2 - 2√2 + 1)/4

= ((√2 – 1)/2)2

⇒ x – (√2 + 1)/2 = ±(√2 – 1)/2 (Taking square root on both sides)

⇒ x – (√2 + 1)/2 + (√2 – 1)/2 or x – (√2 + 1)/2 = -(√2 – 1)/2

⇒ x = (√2 + 1)/2 + (√2 – 1)/2 or x = (√2 + 1)/2 – (√2 – 1)/2

⇒ x = (2√2)/2 = √2 or x = 2/2 = 1

Hence, √2 and 1 are the roots of the given equation.

14. √2x2 – 3x - 2√2 = 0

Solution

√2x2 – 3x - 2√2 = 0

⇒ 2x2 - 3√2x – 4 = 0 (Multiplying both sides by √2)

⇒ 2x2 - 3√2x = 4

⇒ (√2x)2 – 2 × √2x × 3/2 + (3/2)2 = 4 + (3/2)2 [Adding (3/2)2 on both sides]

⇒ (√2x – 3/2)2 = 4 + 9/4 = 25/4 = (5/2)2

⇒ √2x – 3/2 = ± 5/2 (Taking square root on both sides)

⇒ √2x – 3/2 = 5/2 or √2x – 3/2 = -5/2

⇒ √2x = 5/2 + 3/2 = 8/2 = 4 or √2x = -5/2 + 3/2 = -2/2 = -1

⇒ x = 4/√2 = 2√2 or x = -1/√2 = -√2/2 b

Hence, 2√2 and -√2/2 are the roots of the given equation.

15. √3x2 + 10x + 7√3 = 0

Solution

√3x2 + 10x + 7√3 = 0

⇒ 3x2 + 10√3x + 21 = 0 (Multiplying both sides by √3)

⇒ 3x2 + 10√3x = - 21

⇒ (√3x)2 + 2 × √3x × 5 + 52 = -21 + 52 (Adding 52 on both sides)

⇒ (√3x + 5)2 = 21 + 25 = 4 = 22

⇒ √3x + 5 = ±2 (Taking square root on both sides)

⇒ √3x + 5 = 2 or √3x + 5 = - 2

⇒ √3x = -3 or √3x = - 7

⇒ x = -3/√3 = - √3 or x = -7/√3 = -7√3/3

Hence, -√3 and –(7√3)/3 are the roots of the given equation.

16. By using the method of completing the square, show that the equation 2x2 + x + 4 = 0 has no real roots.

Solution

2x2 + x + 4 = 0

⇒ 4x2 + 2x + 8 = 0 (Multiplying both sides by 2)

⇒ 4x2 + 2x = - 8

⇒ (2x)2 + 2 × 2x × (1/2)2 = - 8 + (1/2)2 [Adding (1/2)2 on both sides]

⇒ (2x + 1/2)2 = (- 8 + 1/4) = - (31/4) < 0

But, (2x + 1/2)2 cannot be negative for any real value of x.

So, there is no real value of x satisfying the given equation.

Hence, the given equation has no real roots.