The question I am trying to solve is this:

$4 x^2 - 3 x - 3 = 0$ has roots $p, q$. Find all quadratic equations with roots $p^3$ and $q^3$.

I was able to answer this question by simply finding the roots of the equation using the quadratic formula ($p = -0.5687...$ and $q = 1.3187...$), cubing it, and then plugging it in as $a(x - p)(x - q)$.

This seems to work, however is quite ugly, and is not exact.

The correct answer, according to my textbook, is $a(64 x^2 - 135 x - 27)$, but I can't work out how it got that answer.


Use Vieta's formula to find $\displaystyle p+q=\frac34,pq=-\frac34$

Now, $\displaystyle p^3+q^3=(p+q)^3-3pq(p+q), p^3q^3=(pq)^3$

  • $\begingroup$ Thanks, I will give that a go $\endgroup$ – JAS May 5 '14 at 8:29
  • $\begingroup$ This may seem like a silly question, but how did you get $p^3+q^3=(p+q)^3−3pq(p+q)$ ? I understand how you solve the problem from there, but am not sure how you got to that point $\endgroup$ – JAS May 5 '14 at 8:53
  • $\begingroup$ @JAS, $$(p+q)^3=?$$ $\endgroup$ – lab bhattacharjee May 5 '14 at 11:24
  • $\begingroup$ Thanks, I think I understand how you got this answer now. When you expand $(p+q)^3$, we get an expression which, when you subtract $-3pq(p+q)$ is equal to $p^3+Q^3$ right? $\endgroup$ – JAS May 6 '14 at 7:03

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