x2 + (x – 1)/3 = 0

(iii) x² + ^{(x – 1)}/₃ = 0

Sol. x² + ^{(x – 1)}/₃ = 0

Multiplying the above equation by 3, we get

3x² + x – 1 = 0

Comparing with ax² + bx + c = 0 we have a = 3, b = 1, c = – 1

b² – 4ac = (1)² – 4 (3) (– 1)

= 1 + 12

= 13

By Formula method,

x	=	-  b ± √(b2 – 4ac)
		2a
∴ x	=	-1 ± √13
		2(3)
∴ x	=	-1 ± √13
		6
∴ x	=	-1+√13	or	-1-√13
		6		6