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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
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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