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x2 + x + 1 = 0
(iii) x
2
+ x + 1 = 0
Sol.
x
2
+ x + 1 = 0
Comparing with ax
2
+ bx + c = 0 we have a = 1, b = 1, c = 1
∆
= b
2
– 4ac
= (1)
2
– 4 (1) (1)
= 1 – 4
= – 3
∴ ∆ = – 3
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