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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+37167 
+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+189 
+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.

 

3x2+4x9=2x26x+1x2+10x10=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+210+r22=100r21+r22=120

 

We have found the sum of the square of the roots!

 Aug 24, 2023

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