The number of pairs (a, b) of real numbers, such that whenever α is...

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, $\alpha = {\alpha}^{2} - 2$

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

$⇒ \alpha = - 1$ or $\alpha = 2$

When $\alpha = - 1$ then \left(a, b\right) = \left(2, 1\right)$</p><p>$\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)$</p><p>Hence \left(a, b\right) = \left(- \left(\alpha + \beta\right), \alpha\beta\right)$

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

(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)$</p><p>$= \left(1, - 1\right)$</p><p><strong>(III)</strong>$\alpha = {\alpha}^{2} - 2 = {\beta}^{2} - 2$and$\alpha \ne \beta$</p><p>$⇒ \alpha = - \beta$</p><p>Thus$\alpha = 2, \beta = - 2$</p><p>$\alpha = - 1, \beta = 1$</p><p>Therefore \left(a, b\right) =$\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)$</p><p>Which are \left(2, 1\right), \left(- 4, 4\right), \left(- 1, - 2\right), \left(1, - 1\right), \left(0, - 4\right), \left(0, - 1\right)$