Question

The number of pairs $(a,b)$ of real numbers, such that whenever $\alpha$ is a root of the equation $x^2+ax+b=0,{\alpha }^2-2$ is also a root of this equation, is

• A. $8$
• B. $4$
• C. $6$
• D. $2$

Answer 1

The correct option is C $6$

Let $\alpha ,\beta$ be the roots of the quadratic equation.
Then, $\alpha ={\beta }^2-2$ and $\beta ={\alpha }^2-2$
$\Rightarrow ({\alpha }^2-2{)}^2-2=\alpha$
$\Rightarrow {\alpha }^4-4{\alpha }^2-\alpha +2=0$
$\Rightarrow (\alpha +1)(\alpha -2)({\alpha }^2+\alpha -1)=0$
$\Rightarrow (\alpha ,\beta )=(-1,-1),(-1,1),(2,2),(2,-2),(-1,2)$ and $(\frac{\sqrt{5}-1}{2},-\frac{\sqrt{5}+1}{2})$
Hence there will be $6$ possible values of $(a,b)$.