document.write( "Question 162007: find three consecutive integers such that the sum of the squares of the second and the third exceeds the square of the first by 21. \n" ); document.write( "
Algebra.Com's Answer #119391 by Earlsdon(6294)\"\" \"About 
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Let the three consecutive integers be: x, (x+1), and (x+2)
\n" ); document.write( "\"%28x%2B1%29%5E2+%2B+%28x%2B2%29%5E2+=+x%5E2%2B21\" \"...the sum of the squares of the second and third exceed the square of the first by 21.\" Simplify this and solve for x.
\n" ); document.write( "\"%28x%5E2%2B2x%2B1%29%2B%28x%5E2%2B4x%2B4%29+=+x%5E2%2B21\" Combine like-terms.
\n" ); document.write( "\"2x%5E2%2B6x%2B5+=+x%5E2%2B21\" Subtract \"x%5E2\" from both sides.
\n" ); document.write( "\"x%5E2%2B6x%2B5+=+21\" Subtract 21 from both sides.
\n" ); document.write( "\"x%5E2%2B6x-16+=+0\" Solve this quadratic equation by factoring.
\n" ); document.write( "\"%28x-2%29%28x%2B8%29+=+0\" Apply the zero product rule.
\n" ); document.write( "\"x-2+=+0\" or \"x%2B8+=+0\" so...
\n" ); document.write( "\"x+=+2\" or \"x+=+-8\"
\n" ); document.write( "There are two answers to this problem:
\n" ); document.write( "x = 2
\n" ); document.write( "(x+1) = 3
\n" ); document.write( "(x+2) = 4 and...
\n" ); document.write( "x = -8
\n" ); document.write( "x+1 = -7
\n" ); document.write( "x+2 = -6
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