Quadratic fun

First let us assume the two roots of the equation to be $\alpha$ and $\beta$ .Now we will consider the different cases:

Case 1 : $\alpha=\beta$ or $\alpha$ is a double root.Since $\alpha^2-2$ is also a root we get $\alpha=\alpha^2-2$ or $\alpha=-1$ or $\alpha=2$ .

Using Vieta's formulas we get $a=-2\alpha$ and $b=\alpha^2$ which gives $(a,b)=(2,1) or (-4,4)$

Case 2 : $\alpha=\alpha^2-2$ and $\beta=\beta^2-2$

which gives $(\alpha,\beta)=(-1,2) or (2,-1)$ as discussed in case 1.(Note that (-1,-1) and (2,2) aren't taken because they are already in case 1.)

$\Rightarrow$ $(a,b)=(-(\alpha+\beta),\alpha\beta)=(-1,-2)$

Case 3 : $\alpha=\beta^2-2$ and $\beta=\alpha^2-2$

$\Rightarrow$ $\alpha-\beta=\beta^2-\alpha^2=(\beta-\alpha)(\beta+\alpha)$

as $\alpha \neq \beta$ we get $\alpha+\beta=-1$

also $\alpha+\beta=\beta^2+\alpha^2-4=(\beta+\alpha)^2-2\alpha\beta-4$

which gives $\alpha\beta=-1$ as $\alpha+\beta=-1$

Hence $(a,b)=(-(\alpha+\beta),\alpha\beta)=(1,-1)$

Case 4 : $\alpha^2-2=\beta^2-2=\alpha(or \beta)$ and $\alpha \neq \beta$

$\Rightarrow$ $\alpha^2=\beta^2$

$\Rightarrow$ $\alpha=-\beta$ (as $\alpha \neq \beta$ )

Thus, $(\alpha,\beta)=(2,-2) or (-1,1)$ (using $\alpha^2-2=\alpha$ )

$\Rightarrow$ $(a,b)=(0,-4)$ or $(0,-1)$

Hence, Total number of ordered pairs = $\boxed{6}$ which are $(a,b)=(-4,4),(2,1),(-1,-2),(1,-1),(0,-1),(0,-4)$

For y and y^2-2 to be a root of the equation, (y^2-2)+y=a (sum of roots of an equation) and (y^2-2)*y=b (product of roots of an equation)

For both results we can have 2 and 3 roots(possible values) respectively Thus the no of sets can be 2*3= 6