document.write( "Question 280358: find two consequetive even integers whose product equal 120 \n" ); document.write( "

Algebra.Com's Answer #203782 by jsmallt9(3759) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x = the first even integer. Then x+2 = the second even integer. Their product would be x(x+2). This is 120:
\n" ); document.write( " $\"x%28x%2B2%29+=+120\"$
\n" ); document.write( "Now we solve for x. Start by simplifying the left side:
\n" ); document.write( " $\"x%5E2%2B2x+=+120\"$
\n" ); document.write( "Since this is a quadratic equation we want one side to be zero. So subtract 120 from each side:
\n" ); document.write( " $\"x%5E2%2B2x+-+120+=+0\"$
\n" ); document.write( "Now we factor (or use the Quadratic Formula):
\n" ); document.write( " $\"%28x%2B12%29%28x-10%29+=+0\"$
\n" ); document.write( "By the Zero Product Property we know that one of these factors must be zero:
\n" ); document.write( "x+12 = 0 or x-10 = 0
\n" ); document.write( "Solving these we get:
\n" ); document.write( "x = -12 or x = 10
\n" ); document.write( "That make the second even integer...
\n" ); document.write( "x+2 = -12 + 2 = -10 or x+2 = 10 + 2 = 12

\n" ); document.write( "So there are two pairs of consecutive even integers whose product is 120:
\n" ); document.write( "-12 and -10 or 10 and 12 \n" ); document.write( "

\n" );