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avatar+1420 

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

 Jun 27, 2024
 #1
avatar+1230 
+1

First, let's combine all like terms for x and move all terms to one side. We have that

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

 

Using the quadratic equation, we have that

\(x=\frac{-12\pm \sqrt{12^{2}-4\cdot 1(-4)}}{2\cdot 1}\)

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

 

Now, we can use a handy trick to solve this problem. We have that

\((2\sqrt{10}-6)^2+(-2\sqrt{10}-6)^2\)

 

Expanding and distributing in our values, we have

\(-24\sqrt{10}+76+4\left(\sqrt{10}\right)^{2}+12\sqrt{10}-6\left(-2\sqrt{10}-6\right)\)

 

All the radicals basically cancel out. In the end, we are left with 152. 

 

So 152 is our answer. 

 

Thanks! :)

 Jun 27, 2024
 #2
avatar+14 
+1

Simplifying we get \(x^2 + 12x - 4 = 0\). Using the quadratic formula we find the roots to be \(-6 + 2\sqrt{10}\) and \(-6 - 2\sqrt{10}\). Taking their squares, we get \(36 + 4\sqrt{10} + 40\) and \(36 - 4\sqrt{10} + 40\). Adding them together, we get \(\boxed{152}\)

 Jun 27, 2024
 #3
avatar+129725 
+1

Let the roots be a,b

 

Product of the roots  = ab =  -4   →  2ab = -8

 

Sum of the roots =  a + b = -12     square both sides

 

a^2 + b^2 + 2ab = 144

 

a^2 + b^2 - 8 =144

 

a^2 + b^2 =  152

 

 

cool cool cool

 Jun 27, 2024

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