(vi) x (x – 1) = 1
Sol. x (x – 1) = 1
∴ x2 – x = 1 .... (i)
Third term = ( ½ × coefficient of x)2
= ( ½ × - 1)2
= (- ½ )2
= ¼
Adding ¼ to both sides of (i), we get,
∴ x2 – x + ¼ = 1 + ¼
∴ (x – ½ )2 = 5/4
∴ (x – ½) = ±√5/4
∴ x + ½ = ± √5/2
∴ x = - ½ ± √5/2
∴ x = (- 1 + √5)/2 or (-1 - √5)/2
