The sum of roots of equation x^2+px+q=0 is equal to sum of their squares, then prove that: p(p+1)=2q

Wednesday, January 17, 2018

Let a and b be
the roots of the equation. Hence,

$\displaystyle{{a}}^{{2}}+{a}{p}+{q}={0}$ and

$\displaystyle{{b}}^{{2}}+{p}{b}+{q}={0}$

We know that $\displaystyle{a}+{b}={{a}}^{{2}}+{{b}}^{{2}}$ . Adding the two
equations yield:

$\displaystyle{{a}}^{{2}}+{a}{p}+{q}+{{b}}^{{2}}+{p}{b}+{q}={0}$

Hence,

$\displaystyle{{a}}^{{2}}+{{b}}^{{2}}+{\left({a}+{b}\right)}{p}+{2}{q}={0}$

But $\displaystyle{a}+$
b = a^2 + b^2 $\displaystyle{\quad\text{and}\quad}\ldots$

Spooky Notes