2017/002) Given a cubic eqaution $x^3 + px^2+ qx+r =0$ has 2 roots

 that are equal in value and opposite in sign show that $pq=r$

Proof:

Let the roots be $a,-a,b$
so we have $x^3+px^2+qx + r = (x-a)(x + a)(x-b) = (x^2 - a^2)(x-b) = x^3 - b x^2 -a^2 x + ba^2$
giving $p = -b$, $q=-a^2$, $r = ba^2$
hence $pq = a^2b = r$