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

Consider the equation ${x}^{2}+ax+b=0$x2+ax+b=0

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

Either $\alpha =\beta$α=β or $\alpha \neq \beta$α=β

Case (1) If $\alpha =\beta$α=β, then it is repeated root. Given than ${\alpha }^{2}-2$α2−2 is also root

So, $\alpha ={\alpha }^{2}-2$α=α2−2

$$\Rightarrow (\alpha +1)(\alpha -2)=0$$⇒(α+1)(α−2)=0

$\Rightarrow \alpha =-1$⇒α=−1 or $\alpha =2$α=2

When $\alpha =-1$α=−1 then $(a,b)=(2,1)$(a,b)=(2,1)

$\alpha =2$α=2 then $(a,b)=(-4,4)$(a,b)=(−4,4)

Case (2) If $\alpha \neq \beta$α=β, then four possibilities are there

(I) $\alpha ={\alpha }^{2}-2$α=α2−2 and $\beta ={\beta }^{2}-2$β=β2−2

Here $(\alpha ,\beta )=(2,-1)$(α,β)=(2,−1) or $(-1,2)$(−1,2)

Hence $(a,b)=(-(\alpha +\beta ),\alpha \beta )$(a,b)=(−(α+β),αβ)

$$=(-1,-2)$$=(−1,−2)

(II) $\alpha ={\beta }^{2}-2$α=β2−2 and $\beta ={\alpha }^{2}-2$β=α2−2

Then $\alpha -\beta ={\beta }^{2}-{\alpha }^{2}=(\beta -\alpha )(\beta +\alpha )$α−β=β2−α2=(β−α)(β+α)

Since $\alpha \neq \beta$α=β we get $\alpha +\beta ={\beta }^{2}+{\alpha }^{2}-4$α+β=β2+α2−4

$$\alpha +\beta ={(\alpha +\beta )}^{2}-2\alpha \beta -4$$α+β=(α+β)2−2αβ−4

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

$\alpha \beta =-1$αβ=−1. Therefore $(a,b)=(-(\alpha +\beta ),\alpha \beta )$(a,b)=(−(α+β),αβ)

$$=(1,-1)$$=(1,−1)

(III) $\alpha ={\alpha }^{2}-2={\beta }^{2}-2$α=α2−2=β2−2 and $\alpha \neq \beta$α=β

$$\Rightarrow \alpha =-\beta$$⇒α=−β

Thus $\alpha =2,\beta =-2$α=2,β=−2

$$\alpha =-1,\beta =1$$α=−1,β=1

Therefore $(a,b)=(0,-4)$(a,b)=(0,−4) and $(0,-1)$(0,−1)

(IV) $\beta ={\alpha }^{2}-2={\beta }^{2}-2$β=α2−2=β2−2 and $\alpha \neq \beta$α=β is same as (III)

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

Which are $(2,1),(-4,4),(-1,-2),(1,-1),(0,-4),(0,-1)$(2,1),(−4,4),(−1,−2),(1,−1),(0,−4),(0,−1)