The number of pairs a , b of real numbers, such that whenever α is a root of the equation x 2 + a x + b…

Consider the equation $x^2+ax+b=0$

It has two roots (not necessarily real $\alpha$ and $\beta$)

Either $\alpha =\beta$ or $\alpha \ne \beta$

Case (1) If $\alpha =\beta$, then it is repeated root. Given than ${\alpha }^2-2$ is also root

So, $\alpha ={\alpha }^2-2$

$\Rightarrow \left(\alpha +1\right)\left(\alpha -2\right)=0$

$\Rightarrow \alpha =-1$ or $\alpha =2$

When $\alpha =-1$ then $\left(a,b\right)=\left(2,1\right)$

$\alpha =2$ then $\left(a,b\right)=\left(-4,4\right)$

Case (2) If $\alpha \ne \beta$, then four possibilities are there

(I) $\alpha ={\alpha }^2-2$ and $\beta ={\beta }^2-2$

Here $\left(\alpha ,\beta \right)=\left(2,-1\right)$ or $\left(-1,2\right)$

Hence $\left(a,b\right)=\left(-\left(\alpha +\beta \right),\alpha \beta \right)$

$=\left(-1,-2\right)$

(II) $\alpha ={\beta }^2-2$ and $\beta ={\alpha }^2-2$

Then $\alpha -\beta ={\beta }^2-{\alpha }^2=\left(\beta -\alpha \right)\left(\beta +\alpha \right)$

Since $\alpha \ne \beta$ we get $\alpha +\beta ={\beta }^2+{\alpha }^2-4$

$\alpha +\beta ={\left(\alpha +\beta \right)}^2-2\alpha \beta -4$

Thus $-1=1-2\alpha \beta -4$ which implies

$\alpha \beta =-1$. Therefore $\left(a,b\right)=\left(-\left(\alpha +\beta \right),\alpha \beta \right)$

$=\left(1,-1\right)$

(III) $\alpha ={\alpha }^2-2={\beta }^2-2$ and $\alpha \ne \beta$

$\Rightarrow \alpha =-\beta$

Thus $\alpha =2,\beta =-2$

$\alpha =-1,\beta =1$

Therefore $\left(a,b\right)=\left(0,-4\right)$ and $\left(0,-1\right)$

(IV) $\beta ={\alpha }^2-2={\beta }^2-2$ and $\alpha \ne \beta$ is same as (III)

Therefore we get $6$ pairs of $\left(a,b\right)$

Which are $\left(2,1\right),\left(-4,4\right),\left(-1,-2\right),\left(1,-1\right),\left(0,-4\right),\left(0,-1\right)$