1.

Show that x^2+x+1=0 has real roots

Answer»

X^2 + x + 1 = 0

On COMPARING given equation with general equation, we get

a = 1

b = 1

c = 1

Discriminent = b^2 - 4ac

= (1)^2 - 4(1)(1)

= 1 - 4

= - 3

Discriminent is negative.

So the given equation has no real roots.


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