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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?

 Aug 23, 2023
 #1
avatar+36926 
+1

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 

 Aug 23, 2023
 #2
avatar+177 
+1

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!

 Aug 24, 2023

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