$p$ and $q$ are solutions of the equation $x^2 +p*x+ q=0$. Find values $p$ and $q$.

Question

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?

Answer 1

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 $\,p=q=-.5$