document.write( "Question 231946: Find two consecutive integers, such that the sum of their squares is 321 \n" ); document.write( "
Algebra.Com's Answer #171510 by Alan3354(69443)\"\" \"About 
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Find two consecutive integers, such that the sum of their squares is 321
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\n" ); document.write( "x^2 + (x+1)^2 = 321
\n" ); document.write( "2x^2 + 2x + 1 = 321
\n" ); document.write( "2x^2 + 2x - 320 = 0
\n" ); document.write( "x^2 + x - 160 = 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 \"1x%5E2%2B1x%2B-160+=+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=%281%29%5E2-4%2A1%2A-160=641\".
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\n" ); document.write( " Discriminant d=641 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28-1%2B-sqrt%28+641+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%281%29%2Bsqrt%28+641+%29%29%2F2%5C1+=+12.1589889011722\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%281%29-sqrt%28+641+%29%29%2F2%5C1+=+-13.1589889011722\"
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\n" ); document.write( " Quadratic expression \"1x%5E2%2B1x%2B-160\" can be factored:
\n" ); document.write( " \"1x%5E2%2B1x%2B-160+=+%28x-12.1589889011722%29%2A%28x--13.1589889011722%29\"
\n" ); document.write( " Again, the answer is: 12.1589889011722, -13.1589889011722.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-160+%29\"

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\n" ); document.write( "No integer solution.
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