Find the sum of the squares of the roots of $2x^2+4x-1=x^2-8x+3$.

Answer 1

First, we gather all of our terms on one side and set our quadratic equal to 0: $$
x^2-12x+2=0
$$Now, let's call our roots $r_1$ and $r_2$. By Vieta's formulas, we know that the sum of the roots is $r_1+r_2=12$, so we want to find $r_1^2+r_2^2$.

We can use the sum of squares factorization: \begin{align*}
r_1^2+r_2^2&=(r_1+r_2)^2-2r_1r_2\\ &=12^2-2(-12)\\ &=144+24\\ &=\boxed{168}
\end{align*}