document.write( "Question 81697: Geometry. The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?\r
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Algebra.Com's Answer #58525 by ankor@dixie-net.com(22740)\"\" \"About 
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The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?
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\n" ); document.write( "Let x = the length
\n" ); document.write( "Then (x-1) = the width
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\n" ); document.write( "Using pythag.
\n" ); document.write( "x^2 + (x-1)^2 = 4^2
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\n" ); document.write( "x^2 + (x^2 - 2x + 1) = 16: FOILed (x-1)(x-1)
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\n" ); document.write( "2x^2 - 2x + 1 - 16 = 0; subtracted 16 from both sides
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\n" ); document.write( "2x^2 - 2x - 15 = 0; our old friend, the quadratic equation
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\n" ); document.write( "Use the quadratic equation: a=2; b=-2; c=-15
\n" ); document.write( "\"x+=+%28-%28-2%29+%2B-+sqrt%28+-2%5E2+-+4+%2A+2+%2A+-15+%29%29%2F%282%2A2%29+\"
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\n" ); document.write( "\"x+=+%28%2B2+%2B-+sqrt%28+4+%2B+120+%29%29%2F%284%29+\"
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\n" ); document.write( "Do the math here; you should get about: x = -2.28 and x = +3.28
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\n" ); document.write( "Use the positive solution for x: 3.28 is the length
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\n" ); document.write( "Check using pythag and a good calc:
\n" ); document.write( "3.28^2 + 2.28^2 = 15.9 ~ 16 which is 4^2 \n" ); document.write( "
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