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

 Jun 8, 2024
 #1
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Combining all like terms and moving all terms to one side, we get 

\(x^{2}+12x-4=0\)

 

The two roots of this are

\(x=2\sqrt{10}-6\\ x=-2\sqrt{10}-6\)

 

Squaring them we get 

\(4\left(\sqrt{10}\right)^{2}-12\sqrt{10}-6\left(2\sqrt{10}-6\right)+\left(-2\sqrt{10}-6\right)^{2}\)

\(4\cdot 10-12\sqrt{10}-6\left(2\sqrt{10}-6\right)+\left(-2\sqrt{10}-6\right)^{2}\)

 

This simplifies to 152, which is our answer.

 

Thanks! :)

 Jun 8, 2024

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