+1791 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?
+36919 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
+177 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.
\(3x^2 + 4x - 9 = 2x^2 - 6x + 1 \\ x^2 + 10x - 10 = 0\)
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 \(r_1 + r_2 = -10 \text{ and } r_1r_2 = -10\). Our ultimate goal is to find \(r_1^2 + r_2^2\).
\(r_1 + r_2 = -10 \\ r_1^2 + 2r_1r_2 + r_2^2 = 100 \\ r_1^2 + 2 * -10 + r_2^2 = 100 \\ r_1^2 + r_2^2 = 120\)
We have found the sum of the square of the roots!
