Question

The equations $ax^{2}+bx+a=0$ $\left(a, b \in R\right)$ and $x^{3}-2x^{2}+2x-1=0$ have $2$ roots common. Then $a + b$ must be equal to

• A. $1$
• B. $-1$
• C. $0$
• D. None of these

Answer 1

Answer: (C)

We have, $x^{3}-2x^{3}+2x-1=0$
$\Rightarrow \, \left(x-1\right)\left(x^{2}-x+1\right)=0$
$\Rightarrow \, x=1, \frac{1 \pm\sqrt{3}\,i}{2}$
Since coefficients of $ax^{2} + bx + a = 0$ are real,
$\therefore $ The common roots must be complex and conjugate.
Now, $\frac{1+i \sqrt{3}}{2}=-\left(\frac{-1-i\sqrt{3}}{2}\right)=-\omega^{2}$
Substituting $x =-\omega^{2}$ in $ax^{2}+bx+a=0$
$\Rightarrow \, a\omega^{4}-b\omega^{2}+a=0$
$\Rightarrow \, a\omega-b\omega^{2}+a=0$
$\Rightarrow \, a\left(1+\omega\right)-b\omega^{2}=0$
$\Rightarrow \, a\left(-\omega^{2}\right)-b\omega^{2}=0$
$\Rightarrow \, -\omega^{2}\left(a+b\right)=0$
$\Rightarrow \,a+b=0$