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.