Question

The equations $a{x}^2+bx+a=0$ $\left(a,b\in R\right)$ 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 $a{x}^2+bx+a=0$ 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 $a{x}^2+bx+a=0$
$\Rightarrow a{\omega }^4-b{\omega }^2+a=0$
$\Rightarrow a\omega -b{\omega }^2+a=0$
$\Rightarrow a\left(1+\omega \right)-b{\omega }^2=0$
$\Rightarrow a\left(-{\omega }^2\right)-b{\omega }^2=0$
$\Rightarrow -{\omega }^2\left(a+b\right)=0$
$\Rightarrow a+b=0$