Use sum and product of roots to prove type of roots. Suppose $α$ and $β$ are the roots of $x^2-px+q$. Then
$(a)$ Find the quadratic equation whose roots are $α^2+β^{-2}$ and $β^2+α^{-2}$.
$(b)$ Prove that if $(p,q) \in \mathbb{R}$ then this equation will have equal roots if $p = 0$ and $p^2=4q$.
I have tried to solve part (b) several times but I seem to be heading nowhere…there is a lot of algebra involved and I seem to get lost in the forest every time I attempt it.
Note that: "this equation"in (b) refers to the quadratic equation derived in (a)
 A: For part a we have that
$(a)$ Find the quadratic equation whose roots are $α^2+β^{-2}$ and $β^2+α^{-2}$


*

*$p'=α^2+β^{-2}+β^2+α^{-2}=α^2+β^2+\frac{α^2+β^2}{α^2\beta^2}=p^2-2q+\frac{p^2-2q}{q^2}$

*$q'=(α^2+β^{-2})(β^2+α^{-2})=α^2β^2+2+\frac{1}{α^2\beta^2}=q^2+2+\frac1{q^2}$


then
$$x^2-(p^2-2q+\frac{p^2-2q}{q})x+q^2+2+\frac1{q^2}=0$$
and for part (b) we need $p'^2=4q'$ that is
$$p'^2=\left(p^2-2q+\frac{p^2-2q}{q^2}\right)^2=(p^2-2q)^2\left(1+\frac1{q^2}\right)^2=(p^2-2q)^2\left(1+\frac2{q^2}+\frac1{q^4}\right)$$
$$4q'=4\left(q^2+2+\frac1{q^2}\right)=4q^2\left(1+\frac2{q^2}+\frac1{q^4}\right)$$
which implies
$$(p^2-2q)^2=4q^2\iff(p^2-2q-2q)(p^2-2q+2q)=0 \iff p^2=4q \,\lor\, p=0$$
A: I assume that "this equation"in (b) refers to the equation derived in (a). 
Then if $\alpha ^2+\beta^{-2}=\alpha ^{-2}+\beta ^2$ then $$(\alpha+\beta)(\alpha-\beta)=\frac {(\beta-\alpha)(\beta+\alpha)}{\beta^2\alpha^2}$$
Since we have $\alpha\beta \neq 0$, if these expressions are non zero, they have opposite sign. Therefore we have $\alpha=\pm \beta$
I think you need $p=0$ or (not and) $p^2=4q$ in (b) to make sense of the question.
I did this without completing part (a), which might help to achieve an alternative solution. This is a sketch, so you will have some details to fill in.
