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Given that $p,q$ are roots of the equation $x^2+p*x+q=0$. Find values $p$ and $q$. One method of finding a solution is using Viète’s Theorem. So, $p+q=-p$ and $p*q=q$ and there are two solutions $p=0, q=0$ and $p=1, q=-2.$

Another method is substituting $p$ and $q$ into the equation and solve the system of equations:

$p^2+p^2+q=0$, $q^2+p*q+q=0$

There are three solutions for the system:

  1. $p=0, q=0,$
  2. $p=1, q=-2,$
  3. $p=-.5, q=-.5.$

Why does #3 here, p = q = -.5, does not satisfy the Viète’s Theorem?

TwoTea
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Mar 03 '23 at 15:56
  • Viete's theorem assumes the roots of the polynomial are exactly $p$ and $q$. But when $p = q$, the polynomial could have both $p$ and $q$ as one root and some other number as the second. – eyeballfrog Mar 03 '23 at 16:42
  • If $x^2+p*x +q=0$, then how is "The obvious solution is $p=0, q=0$."? Please explain the reasoning of how you got this conclusion. – Somos Mar 03 '23 at 17:37
  • @Somos The question is asking for $p$ and $q$ such that they are roots of $x^2 + p x + q$. $p=q=0$ can be found by inspection. – eyeballfrog Mar 03 '23 at 20:23

1 Answers1

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The problem is that if the equation $\,x^2 + px + q = 0\,$ has the solutions $\,x = p = q,\,$ then

$$ (x - p)(x - q) = x^2 - (p+q)x + (pq) = x^2 - 2px + p^2. $$

But this equation is also $\,x^2 + px + q = 0\,$ which implies $\, -2p = p \,$ and $\, p^2 = q\,$ and hence $\, p = q = 0.$

Your two equations

$p^2+p^2+q=0$, $q^2+p*q+q=0$

are actually the same equation if $\,p=q\,$ and in that case, the equations are both equal to

$$p^2 + p^2 + p = p(1 + 2p) = 0$$

with two roots $p = 0$ and $p = -.5$. However,

$$ x^2 -.5x -.5 = (x+.5)(x-1). $$

In both equations "Viète’s Theorem" holds true.

What you do not have is a quadratic with two roots $\,p=q=-.5$

Somos
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