The quadratic equation $3x^2+4x-9 = 2x^2-6x+1$ has two real roots. What is the sum of the squares of these roots?
3x^2+4x-9 = 2x^2-6x+1 putting all of the terms on the left side results in:
x^2 + 10x -10 = 0
Use Quadratic Formula with a = 1 b = 10 c = -10
to find roots -5 - sqrt (35) and -5 + sqrt (35) squaring these:
25 + 10 sqrt35 + 35 25 - 10 sqrt 35 + 35
summed = 120
ElectricPavlov's method is perfectly valid, but there is a method that avoids having to do arithmetic with square roots, which can become quite messy. I mostly thank Cphill for presenting Vieta's formula in a previous answer. Make sure the quadratic equation is in standard form first.
3x2+4x−9=2x2−6x+1x2+10x−10=0
By Vieta's formula, we can deduce the product of the roots is the constant term, and the sum of the roots is the opposite of the coefficient of the x-term. This means that r1+r2=−10 and r1r2=−10. Our ultimate goal is to find r21+r22.
r1+r2=−10r21+2r1r2+r22=100r21+2∗−10+r22=100r21+r22=120
We have found the sum of the square of the roots!