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

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 10A Exercise 10C Exercise 10D Exercise 10E Exercise 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 3² on 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

x² + 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

4x² + 4√3x + 3 = 0

⇒ 4x² + 4√3x = -3

⇒ (2x)² + 2 × 2x × √3 + (√3)²

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

⇒ (2x + √3)² = - 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

2x² + 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. 3x² – x – 2 = 0

Solution

3x² – 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

⇒ 16x² – 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)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.