document.write( "Question 122798: Can anyone help with this problem?\r
\n" ); document.write( "\n" ); document.write( "Find two consecutive odd positive integers such that the twice the square of the larger minus the square of the smaller is 41 (check your answer)\r
\n" ); document.write( "\n" ); document.write( "larger = x
\n" ); document.write( "smaller = y\r
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Algebra.Com's Answer #90161 by oscargut(2103)\"\" \"About 
You can put this solution on YOUR website!
If x is the larger and y the smallest then y=x-2 because are consecutive odds
\n" ); document.write( "then 2x^2-(x-2)^2=41
\n" ); document.write( " then x^2+4x-4=41 then x^2+4x-45=0 then x=5 or x=-9
\n" ); document.write( " so x=5 and y=x-2=3\r
\n" ); document.write( "\n" ); document.write( "The numbers are 5 and 3\r
\n" ); document.write( "\n" ); document.write( "Check:\r
\n" ); document.write( "\n" ); document.write( "twice the square of the larger is 2(5^2)=50
\n" ); document.write( "the square of the smaller is 3^2=9\r
\n" ); document.write( "\n" ); document.write( "so the twice the square of the larger minus the square of the smaller is 41
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