<s>[INST] Forming a quadratic equation with the given roots Question: If $\frac{p^2}{q}$ and $\frac{q^2}{p}$ are the roots of the equation $2x^2+7x-4=0$, find the equation whose roots are $p$ and $q$($p+q$ is real).
My attempt: The required equation is $x^2-(p+q)x+pq$
$(\frac{p^2}{q})(\frac{q^2}{p})=\frac{-4}{2}$
$pq=-2$
$\frac{p^2}{q}+\frac{q^2}{p}=\frac{-7}{2}$
$\frac{p^3+q^3}{pq}=\frac{-7}{2}$
$p^3+q^3=7$
$(p+q)^3-3pq(p+q)=7$
$(p+q)^3+6(p+q)=7$
My problem:I am unable to solve this equation. I think this one is a cubic equation and i do not know how to solve them. Moreover it is not in our syllabus so i there there must be another approach to find $p+q$
 [/INST] $$\frac{p^2}{q}+\frac{q^2}{p}=-\frac{7}{2}$$ and $pq=-2$.
Thus, $$p^3+q^3=7$$ or
$$(p+q)((p+q)^2+6)=7$$ or
$$(p+q)^3+6(p+q)-7=0,$$
which gives $p+q=1$ and we get the answer:
$$x^2-x-2=0.$$
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