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

 Apr 2, 2024
 #1
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Find the sum of the squares of the roots of $2x^2+4x-1=x^2-8x+3$.  

 

                                           2x2 + 4x – 1  =  x2 – 8x + 3  

Combine like terms               x2 + 12x – 4  =  0  

 

Complete the square            x2 + 12x + 36  =  40  

                                             (x + 6)2  =  40  

                                              x + 6  =  +sqrt(40)  

                                                    x  =  +sqrt(40) – 6  

 

Square each root      [ +sqrt(40) – 6 ]2  =  40 – 12sqrt(40) + 36      (eq 1)   

                                  [ –sqrt(40) – 6 ]2  =  40 + 12sqrt(40) + 36      (eq 2)   

 

Add (eq 1) & (eq 2)                                    80                      + 72   

 

                              Sum of squares  =  152    

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 Apr 2, 2024
 #2
avatar+6 
+1

The given equation is:
2x2+4x−1=x2−8x+3

Rearranging terms, we get:
2x2+4x−1−x2+8x−3=0
x2+12x−4=0
Now, the sum of the squares of the roots of this equation can be found using the formula:

Sum of the squares of the roots=(Sum of the roots)2−2(Product of the roots)
 

For the equation $x^2 + bx + c = 0$, the sum of the roots is $-b$ and the product of the roots is $c$. Therefore, for our equation $x^2 + 12x - 4 = 0$, the sum of the roots is $-12$ and the product of the roots is $-4$.

Plugging these values into the formula, we get: Sum of the squares of the roots=(−(−12*12))−2(−4)
=(12*12)+8
=144+8
=152

So, the sum of the squares of the roots of the given equation is 152.

 Apr 3, 2024

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