Question

EXERCISE 4.3
1. Find the roots of the following quadratic equations, if they exist, by the method of
completing the
square:
(1) 2x2 - 7x +3=0
(ii) 2x2 + x-4=0
(iii) 4x²+43x+3 = 0
(iv) 2x + x +4=0​

Answer 1

Answer:

i) 2x2 – 7x + 3 = 0

⇒ 2x2 – 7x = - 3

On dividing both sides of the equation by 2, we get

⇒ x2 – 7x/2 = -3/2

⇒ x2 – 2 × x × 7/4 = -3/2

On adding (7/4)2 to both sides of equation, we get

⇒ (x)2 - 2 × x × 7/4 + (7/4)2 = (7/4)2 - 3/2

⇒ (x - 7/4)2 = 49/16 - 3/2

⇒ (x - 7/4)2 = 25/16

⇒ (x - 7/4) = ± 5/4

⇒ x = 7/4 ± 5/4

⇒ x = 7/4 + 5/4 or x = 7/4 - 5/4

⇒ x = 12/4 or x = 2/4

⇒ x = 3 or 1/2

(ii) 2x2 + x – 4 = 0

⇒ 2x2 + x = 4

On dividing both sides of the equation, we get

⇒ x2 + x/2 = 2

On adding (1/4)2 to both sides of the equation, we get

⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2

⇒ (x + 1/4)2 = 33/16

⇒ x + 1/4 = ± √33/4

⇒ x = ± √33/4 - 1/4

⇒ x = ± √33-1/4

⇒ x = √33-1/4 or x = -√33-1/4

(iii) 4x2 + 4√3x + 3 = 0

⇒ (2x)2 + 2 × 2x × √3 + (√3)2 = 0

⇒ (2x + √3)2 = 0

⇒ (2x + √3) = 0 and (2x + √3) = 0

⇒ x = -√3/2 or x = -√3/2

(iv) 2x2 + x + 4 = 0

⇒ 2x2 + x = -4

On dividing both sides of the equation, we get

⇒ x2 + 1/2x = 2

⇒ x2 + 2 × x × 1/4 = -2

On adding (1/4)2 to both sides of the equation, we get

⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 - 2

⇒ (x + 1/4)2 = 1/16 - 2

⇒ (x + 1/4)2 = -31/16

However, the square of number cannot be negative.

Therefore, there is no real root for the given equation