3
$\begingroup$

How to prove that

$$ \begin{cases} x_1 + x_2 + x_3 & = 0 \\ x_1x_2 + x_2x_3 + x_3x_1 & = p \\ x_1x_2x_3 & = -q \\ x_1 & = 1/x_2 + 1/x_3 \end{cases} $$

implies

$$ q^3 + pq + q = 0\,\,? $$

$\endgroup$
  • 1
    $\begingroup$ Have you tried subbing in your values for p and q and then using the given identities? $\endgroup$ – creilly May 28 '14 at 18:45
  • 1
    $\begingroup$ First three are Vieta's formula. Not sure about last one. $\endgroup$ – Kaster May 28 '14 at 19:02
1
$\begingroup$

Denote the equations by $(1),\ldots ,(4)$. Then $(4)$ says $x_1x_2x_3=x_2+x_3$ and $(3)$ says $x_1x_2x_3=-q$. This gives $x_3=-x_2-q$. Substitute this into $(1)$. This gives $x_1=q$. Then $q\cdot (2)-(3)$ gives $-q(p+q^2+1)=0$, or $$q^3+pq+q=0.$$

$\endgroup$
1
$\begingroup$

According to the first three equations, $x_1, x_2, x_3$ are solutions of $$x^3 +px + q = 0$$

The fourth equation can be translated into $x_1x_2x_3 = x_3 + x_2$, i.e. $-q = -x_1$, then $x_1 = q$,

Conclude by noting that $q$ solves $x^3 +px + q = 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.