document.write( "Question 330195: Find three consecutive multiples of 5 such that the sum of the squares of the first two is 125 greater than the square of the third.\r
\n" ); document.write( "\n" ); document.write( "This is what I did, but it's wrong.\r
\n" ); document.write( "\n" ); document.write( " $\"%285N%29%5E2+%2B+%285N%2B5%29%5E2+-+125+=+%285N%2B10%29%5E2\"$
\n" ); document.write( "25N + 25N + 25 - 125 = 25N + 100
\n" ); document.write( "25N - 100 = 100
\n" ); document.write( "25N = 200
\n" ); document.write( "N = 8 \n" ); document.write( "

Algebra.Com's Answer #236634 by Alan3354(69443) $\"\"$ $\"About$

You can put this solution on YOUR website!
Find three consecutive multiples of 5 such that the sum of the squares of the first two is 125 greater than the square of the third.\r
\n" ); document.write( "\n" ); document.write( "This is what I did, but it's wrong.\r
\n" ); document.write( "\n" ); document.write( " $\"%285N%29%5E2+%2B+%285N%2B5%29%5E2+-+125+=+%285N%2B10%29%5E2\"$
\n" ); document.write( "This is ok, but you didn't square any terms.
\n" ); document.write( " $\"25n%5E2+%2B+25n%5E2+%2B+50n+%2B+25+-+125+=+25n%5E2+%2B+100n+%2B+100\"$
\n" ); document.write( " $\"50n%5E2+%2B+50n+-100+=+25n%5E2+%2B+100n+%2B+100\"$
\n" ); document.write( " $\"25n%5E2+-+50n+-+200+=+0\"$
\n" ); document.write( " $\"n%5E2+-+2n+-+8+=+0\"$
\n" ); document.write( "(n-4)*(n+2) = 0
\n" ); document.write( "n = 4
\n" ); document.write( "-----------\r
\n" ); document.write( "\n" ); document.write( "25N + 25N + 25 - 125 = 25N + 100
\n" ); document.write( "25N - 100 = 100
\n" ); document.write( "25N = 200
\n" ); document.write( "N = 8\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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