document.write( "Question 666507: Direction are complete the square.\r
\n" ); document.write( "\n" ); document.write( "x^2+2x=5\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "I get to x^2+2x-5=0
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Algebra.Com's Answer #414554 by Leaf W.(135)\"\" \"About 
You can put this solution on YOUR website!
\"x%5E2+%2B+2x+=+5\"
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\n" ); document.write( "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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\n" ); document.write( "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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\n" ); document.write( "Square this number: \"1%5E2\" = 1
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\n" ); document.write( "Add the result to both sides of the equation: \"x%5E2+%2B+2x+%2B+1+=+5+%2B+1\"
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\n" ); document.write( "Simplify the right side of the equation: \"x%5E2+%2B+2x+%2B+1+=+6\"
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\n" ); document.write( "Factor the left side of the equation: \"%28x+%2B+1%29%5E2+=+6\"
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\n" ); document.write( "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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\n" ); document.write( "Isolate the x by subtracting 1 from both sides: x = -1 ± \"sqrt%286%29\"
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\n" ); document.write( "Your final answer is x = -1 ± \"sqrt%286%29\".
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