The numbers of pairs (a, b) of real numbers, such that whenever $\alpha$ is a root of the equation x² + ax + b = 0, $\alpha$² $-$ 2 is also a root of this equation, is :

Consider the equation x² + ax + b = 0<br><br>If has two roots (not necessarily real $\alpha$ &amp; $\beta$)<br><br>Either $\alpha$ = $\beta$ or $\alpha$ $\ne$ $\beta$<br><br>Case (1) If $\alpha$ = $\beta$, then it is repeated root. Given that $\alpha$² $-$ 2 is also a root<br><br>So, $\alpha$ = $\alpha$² $-$ 2 $\Rightarrow$ ($\alpha$ + 1)($\alpha$ $-$ 2) = 0<br><br>$\Rightarrow$ $\alpha$ = $-$1 or $\alpha$ = 2<br><br>When $\alpha$ = $-$1 then (a, b) = (2, 1)<br><br>$\alpha$ = 2 then (a, b) = ($-$4, 4)<br><br>Case (2) If $\alpha$ $\ne$ $\beta$<br><br>Then<br><br>(I) $\alpha$ = $\alpha$² $-$ 2 and $\beta$ = $\beta$² $-$ 2<br><br>Hence, (a, b) = ($-$($\alpha$ + $\beta$), $\alpha$$\beta$)<br><br>($-$1, $-$2)<br><br>(II) $\alpha$ = $\beta$² $-$ 2 and $\beta$ = $\alpha$² $-$ 2<br><br>Then $\alpha$ $-$ $\beta$ = $\beta$² $-$ $\alpha$² = ($\beta$ $-$ $\alpha$) ($\beta$ + $\alpha$)<br><br>Since $\alpha$ $\ne$ $\beta$ we get $\alpha$ + $\beta$ = $\beta$² + $\alpha$² $-$ 4<br><br>$\alpha$ + $\beta$ = ($\alpha$ + $\beta$)² $-$ 2$\alpha$$\beta$ $-$ 4<br><br>Thus $-$1 = 1 $-$2 $\alpha$$\beta$ $-$ 4 which implies<br><br>$\alpha$$\beta$ = $-$1 Therefore (a, b) = ($-$($\alpha$ + $\beta$), $\alpha$$\beta$)<br><br>= (1, $-$1)<br><br>(III) $\alpha$ = $\alpha$² $-$ 2 = $\beta$² $-$ 2 and $\alpha$ $\ne$ $\beta$<br><br>$\Rightarrow$ $\alpha$ = $-$ $\beta$<br><br>Thus $\alpha$ = 2, $\beta$ = $-$2<br><br>$\alpha$ = $-$1, $\beta$ = 1<br><br>Therefore (a, b) = (0, $-$4) &amp; (0, 1)<br><br>(IV) $\beta$ = $\alpha$² $-$ 2 = $\beta$² $-$ 2 and $\alpha$ $\ne$ $\beta$ is same as (III) Therefore we get 6 pairs of (a, b)<br><br>Which are (2, 1), ($-$4, 4), ($-$1, $-$2), (1, $-$1), (0, $-$4)<br><br>Option (a)