document.write( "Question 228601: One number is three more than a second number. The sum of their squares is 37 more than the product of the numbers. Find the numbers. \n" ); document.write( "

Algebra.Com's Answer #169646 by rapaljer(4671) $\"\"$ $\"About$

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Let x = second number
\n" ); document.write( "x+3 = first number\r
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\n" ); document.write( "\n" ); document.write( "Sum of squares IS 37 more than the product of the numbers.\r
\n" ); document.write( "\n" ); document.write( "x^2 + (x+3)^2= x(x+3) + 37
\n" ); document.write( "x^2 + x^2 + 6x +9 = x^2 +3x + 37\r
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\n" ); document.write( "\n" ); document.write( "Set equation equal to zero by subtracting x^2 +3x + 37 from each side:
\n" ); document.write( "2x^2 + 6x + 9-x^2 -3x-37=0
\n" ); document.write( "x^2 + 3x - 28=0\r
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\n" ); document.write( "\n" ); document.write( "This just happens to factor!
\n" ); document.write( "(x+7)(x-4)=0\r
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\n" ); document.write( "\n" ); document.write( "There are two solutions:\r
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\n" ); document.write( "\n" ); document.write( "First solution:
\n" ); document.write( "x=-7
\n" ); document.write( "x+3= -4\r
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\n" ); document.write( "\n" ); document.write( "Second solution:
\n" ); document.write( "x=4
\n" ); document.write( "x+3= 7\r
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\n" ); document.write( "\n" ); document.write( "Both solutions do check if you take the time to do it.\r
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\n" ); document.write( "\n" ); document.write( "Dr. Robert J. Rapalje, Retired
\n" ); document.write( "Seminole State College of Florida
\n" ); document.write( "Altamonte Springs Campus \r
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