Exercise 4.3 ka first question

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x² – 7x +3 = 0

(ii) 2x² + x – 4 = 0
(iii) 4x² + 4√3x + 3 = 0

(iv) 2x² + x + 4 = 0

Solutions:

(i) 2x² – 7x + 3 = 0

⇒ 2x² – 7x = – 3

Dividing by 2 on both sides, 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)² = ±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 x = 1/2

(ii) 2x² + x – 4 = 0

⇒ 2x² + x = 4

Dividing both sides of the equation by 2, we get

⇒ x² +x/2 = 2

Now 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

Therefore, either x = √33-1/4 or x = -√33-1/4

(iii) 4x² + 4√3x + 3 = 0

Converting the equation into a²+2ab+b² form, we get,

⇒ (2x)² + 2 × 2x × √3 + (√3)² = 0

⇒ (2x + √3)² = 0

⇒ (2x + √3) = 0 and (2x + √3) = 0

Therefore, either x = -√3/2 or x = -√3/2.

(iv) 2x² + x + 4 = 0

⇒ 2x² + x = -4

Dividing both sides of the equation by 2, we get

⇒ x² + 1/2x = 2

⇒ x² + 2 × x × 1/4 = -2

By 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

As we know, the square of numbers cannot be negative.

Therefore, there is no real root for the given equation, 2x² + x + 4 = 0.