Question 81697
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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Let x = the length
Then (x-1) = the width
:
Using pythag.
x^2 + (x-1)^2 = 4^2 
:
x^2 + (x^2 - 2x + 1) = 16: FOILed (x-1)(x-1)
:
2x^2 - 2x + 1 - 16 = 0; subtracted 16 from both sides
:
2x^2 - 2x - 15 = 0; our old friend, the quadratic equation
:
Use the quadratic equation: a=2; b=-2; c=-15
{{{x = (-(-2) +- sqrt( -2^2 - 4 * 2 * -15 ))/(2*2) }}}
:
{{{x = (+2 +- sqrt( 4 + 120 ))/(4) }}}
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Do the math here; you should get about:  x = -2.28 and x = +3.28
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Use the positive solution for x: 3.28 is the length
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Check using pythag and a good calc:
3.28^2 + 2.28^2 = 15.9 ~ 16 which is 4^2