document.write( "Question 144460: Twice the square of an integer is 21 more than eleven times the integer. Find the integer. \n" ); document.write( "
Algebra.Com's Answer #105197 by Alan3354(69443)\"\" \"About 
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\"2I%5E2+=+11I+%2B+21\"
\n" ); document.write( "\"2I%5E2+-11I+-21+=+0\"
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"2x%5E2%2B-11x%2B-21+=+0\") has the following solutons:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%28-11%29%5E2-4%2A2%2A-21=289\".
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\n" ); document.write( " Discriminant d=289 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28--11%2B-sqrt%28+289+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%28-11%29%2Bsqrt%28+289+%29%29%2F2%5C2+=+7\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%28-11%29-sqrt%28+289+%29%29%2F2%5C2+=+-1.5\"
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\n" ); document.write( " Quadratic expression \"2x%5E2%2B-11x%2B-21\" can be factored:
\n" ); document.write( " \"2x%5E2%2B-11x%2B-21+=+%28x-7%29%2A%28x--1.5%29\"
\n" ); document.write( " Again, the answer is: 7, -1.5.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-11%2Ax%2B-21+%29\"

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\n" ); document.write( "\n" ); document.write( "-1.5 is not an integer, so the answer is 7.\r
\n" ); document.write( "\n" ); document.write( "2 times 7 times 7 = 11 times 7 + 21
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