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Eli Kline
Mathematics
13 October, 15:33
Solve for x in the equation 2x^2+3x-7=x^2+5x+39
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Dereon
13 October, 15:57
0
Subtract x^2 from both sides
x^2 + 3x - 7 = 5x + 39
Subtract 5x from both sides
x^2 - 2x - 7 = 39
Add 7 to both sides
x^2 - 2x = 46
Complete the square by adding (b/2) ^2 to both sides, b = (-2)
(-2/2) = - 1, then square that (-1) ^2 = 1
x^2 - 2x + 1 = 46 + 1
Simplify the expression by factoring
(x - 1) ^2 = 47
Take square root on each side
x - 1 = (sqrt (47))
Solve for x
x = 1 + (sqrt (47))
Since 47 is prime, 47 cannot be broken down by the square root and this is the answer to your problem.
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