The number of pairs (a, b) of real numbers su | vedclass

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(A) Let the roots of $x^{2}+ax+b=0$ be $\alpha$ and $\beta$.If $\alpha$ is a root,then $\alpha^{2}-2$ must also be a root.Case $1$: $\alpha = \beta$.

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$6$

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(A) Let the roots of $x^{2}+ax+b=0$ be $\alpha$ and $\beta$.If $\alpha$ is a root,then $\alpha^{2}-2$ must also be a root.Case $1$: $\alpha = \beta$. Then $\alpha = \alpha^{2}-2$,so $\alpha^{2}-\alpha-2=0$,which gives $\alpha=2$ or $\alpha=-1$.If $\alpha=2$,then $x^{2}-4x+4=0$,so $(a, b) = (-4, 4)$.If $\alpha=-1$,then $x^{2}+2x+1=0$,so $(a, b) = (2, 1)$.Case $2$: $\alpha 
eq \beta$. The set of roots $S = \{\alpha, \beta\}$ must be mapped to itself by $f(x) = x^{2}-2$.Subcase $2.1$: $f(\alpha)=\alpha$ and $f(\beta)=\beta$. This leads to $\alpha, \beta \in \{2, -1\}$. Since $\alpha 
eq \beta$,we have $\{\alpha, \beta\} = \{2, -1\}$. Then $a = -(\alpha+\beta) = -1$ and $b = \alpha\beta = -2$. So $(a, b) = (-1, -2)$.Subcase $2.2$: $f(\alpha)=\beta$ and $f(\beta)=\alpha$. Then $\alpha^{2}-2=\beta$ and $\beta^{2}-2=\alpha$. Subtracting gives $\alpha^{2}-\beta^{2} = \beta-\alpha$,so $(\alpha-\beta)(\alpha+\beta+1)=0$. Since $\alpha 
eq \beta$,$\alpha+\beta = -1$. Also $\alpha^{2}+\beta^{2}-4 = \alpha+\beta = -1$,so $(\alpha+\beta)^{2}-2\alpha\beta = 3$,which gives $1-2\alpha\beta=3$,so $\alpha\beta=-1$. Thus $a = -(\alpha+\beta) = 1$ and $b = \alpha\beta = -1$. So $(a, b) = (1, -1)$.Subcase $2.3$: $f(\alpha)=f(\beta)=\alpha$ (or $\beta$). If $f(\alpha)=f(\beta)=\alpha$,then $\alpha^{2}-2=\alpha$ and $\beta^{2}-2=\alpha$. Thus $\alpha \in \{2, -1\}$. If $\alpha=2$,then $\beta^{2}-2=2$ $\Rightarrow \beta^{2}=4$ $\Rightarrow \beta=-2$ (since $\beta 
eq \alpha$). Then $a = -(2-2)=0$ and $b = 2(-2)=-4$. So $(a, b) = (0, -4)$. If $\alpha=-1$,then $\beta^{2}-2=-1$ $\Rightarrow \beta^{2}=1$ $\Rightarrow \beta=1$ (since $\beta 
eq \alpha$). Then $a = -(-1+1)=0$ and $b = -1(1)=-1$. So $(a, b) = (0, -1)$.There are $6$ such pairs: $(2, 1), (-4, 4), (-1, -2), (1, -1), (0, -4), (0, -1)$.

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:

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