Question:
If $a, b$ are the roots of the equation $x^{2}+x+1=0$, then $a^{2}+b^{2}=$
(a) 1
(b) 2
(c) −1
(d) 3
Solution:
(c) −1
Given equation: $x^{2}+x+1=0$
Also, $a$ and $b$ are the roots of the given equation.
Sum of the roots $=a+b=\frac{-C \text { oefficient of } x}{C \text { oefficient of } x^{2}}=-\frac{1}{1}=-1$
Product of the roots $=a b=\frac{C \text { onstant term }}{C \text { oefficient of } x^{2}}=\frac{1}{1}=1$
$\therefore(a+b)^{2}=a^{2}+b^{2}+2 a b$
$\Rightarrow(-1)^{2}=a^{2}+b^{2}+2 \times 1$
$\Rightarrow 1-2=a^{2}+b^{2}$
$\Rightarrow a^{2}+b^{2}=-1$