SOLUTION: I would like some help on this problem, please. "You are hanging a rectangular mirror in your room that has a diagonal 15 inches long. The width of the mirror is three inches short

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Question 422166: I would like some help on this problem, please. "You are hanging a rectangular mirror in your room that has a diagonal 15 inches long. The width of the mirror is three inches shorter than it's length. What are the dimensions of the mirror?"
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
You are hanging a rectangular mirror in your room that has a diagonal 15 inches long. The width of the mirror is three inches shorter than it's length. What are the dimensions of the mirror?
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Let length be "x"
Then width is "x-3"
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Solve:
x^2 + (x-3)^2 = 15^2
x^2 + x^2-6x+9 = 225
2x^2 - 6x - 116 = 0
x^2 - 3x - 58 = 0
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x = [3 +- sqrt(9 - 4*-58)]/2
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Positive solution:
x = [3 + sqrt(241)]/2 (length)
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x -3 = [3 + sqrt(241)-6]/2
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x - 3 = [-3+sqrt(241)]/2 (width)
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Cheers,
Stan H.
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