SOLUTION: Direction are complete the square. x^2+2x=5 Am I suppose to set this to zero? When I do, I don't know what would make the sum of 2 and give me -5. I get to x^2+2x-5=0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Direction are complete the square. x^2+2x=5 Am I suppose to set this to zero? When I do, I don't know what would make the sum of 2 and give me -5. I get to x^2+2x-5=0       Log On


   



Question 666507: Direction are complete the square.
x^2+2x=5
Am I suppose to set this to zero? When I do, I don't know what would make the sum of 2 and give me -5.
I get to x^2+2x-5=0

Answer by Leaf W.(135) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+2x+=+5
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No, you are not supposed to set this equal to zero when you complete the square; that is for quadratic formula. It is in perfect starting format the way it is -- the coefficient of x%5E2 (number it is multiplied; for example, the coefficient of x in 2x is 2) is 1, which is exactly what you want. Also, the two x terms (x^2 and 2x) are on the left of the equation, and the 5 is on the right, which is also correct.
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Divide the coefficient of the "x" term (in this case, the x term would be 2x, so its coefficient would be the number before the x, 2) in half. This would give you 2%2F2, or 1.
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Square this number: 1%5E2 = 1
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Add the result to both sides of the equation: x%5E2+%2B+2x+%2B+1+=+5+%2B+1
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Simplify the right side of the equation: x%5E2+%2B+2x+%2B+1+=+6
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Factor the left side of the equation: %28x+%2B+1%29%5E2+=+6
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Square root both sides. Remember that the right side could be either the positive or the negative square root: x + 1 = ±sqrt%286%29
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Isolate the x by subtracting 1 from both sides: x = -1 ± sqrt%286%29
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Your final answer is x = -1 ± sqrt%286%29.