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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 $p+q$

MrAP
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2 Answers2

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Hint. You are on the right track. Note that $$ x^3+6x-7=(x-1)(x^2+x+7).$$

Robert Z
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  • Can you please explain how you factorized the L.H.S? – MrAP Jun 16 '17 at 11:52
  • @MrAP Note that $1$ is a solution, then divide $x^3+6x-7$ by $x-1$. – Robert Z Jun 16 '17 at 11:53
  • Do you have a general method to factorize this? I don't feel good about using hit and trial method to find a factor of the L.H.S. – MrAP Jun 16 '17 at 12:00
  • Without knowing any explicit solution you have to use a general formula https://en.wikipedia.org/wiki/Cubic_function#General_formula – Robert Z Jun 16 '17 at 13:01
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$$\frac{p^2}{q}+\frac{q^2}{p}=-\frac{7}{2}$$ 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: $$x^2-x-2=0.$$

Michael Rozenberg
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