(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2x2 + 7x + 5√2 = 0
(iv) 2x2 – x + 1/8 = 0
(v) 100x2 – 20x + 1 = 0
Solution:
(i) Given, x2 – 3x – 10 = 0
Taking L.H.S.,
⇒ x2 – 5x + 2x – 10
⇒ x(x – 5) + 2(x – 5)
⇒ (x – 5)(x + 2)
The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0
Therefore, x – 5 = 0 or x + 2 = 0
⇒ x = 5 or x = -2
(ii) Given, 2x2 + x – 6 = 0
Taking L.H.S.,
⇒ 2x2 + 4x – 3x – 6
⇒ 2x(x + 2) – 3(x + 2)
⇒ (x + 2)(2x – 3)
The roots of this equation, 2x2 + x – 6=0 are the values of x for which (x + 2)(2x – 3) = 0
Therefore, x + 2 = 0 or 2x – 3 = 0
⇒ x = -2 or x = 3/2
(iii) √2 x2 + 7x + 5√2 = 0
Taking L.H.S.,
⇒ √2 x2 + 5x + 2x + 5√2
⇒ x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)
The roots of this equation, √2 x2 + 7x + 5√2 = 0 are the values of x for which (√2x + 5)(x + √2) = 0
Therefore, √2x + 5 = 0 or x + √2 = 0
⇒ x = -5/√2 or x = -√2
(iv) 2x2 – x +1/8 = 0
Taking L.H.S.,
= 1/8 (16x2 – 8x + 1)
= 1/8 (16x2 – 4x -4x + 1)
= 1/8 (4x(4x – 1) -1(4x – 1))
= 1/8 (4x – 1)2
The roots of this equation, 2x2 – x + 1/8 = 0, are the values of x for which (4x – 1)2= 0
Therefore, (4x – 1) = 0 or (4x – 1) = 0
⇒ x = 1/4 or x = 1/4
(v) Given, 100x2 – 20x + 1=0
Taking L.H.S.,
= 100x2 – 10x – 10x + 1
= 10x(10x – 1) -1(10x – 1)
= (10x – 1)2
The roots of this equation, 100x2 – 20x + 1=0, are the values of x for which (10x – 1)2= 0
∴ (10x – 1) = 0 or (10x – 1) = 0
⇒ x = 1/10 or x = 1/10