SOLUTION: The width of a rectangle is 16 ft less than the length. The area is 10 ft^2. Find the length and the width.

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Question 1158688: The width of a rectangle is 16 ft less than the length. The area is 10 ft^2. Find the length and the width.
Found 2 solutions by Shin123, josgarithmetic:
Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
Let the length and width of the rectangle be l and w, respectively. We have system%28w=l-16%2Clw=10%29. Substituting l-16 for w in the second equation, we get l%28l-16%29=10. Expanding and rearranging, we get l%5E2-16l-10=0.
Solved by pluggable solver: Completing the Square for Quadratics
To complete the square for the quadratic x%5E2-16%2Ax-10=0, we must first find a square which when expanded, has x2 and -16x in it.
%28x-8%29%5E2 is the square we are looking for. Expanding %28x-8%29%5E2 gets x%5E2-16%2Ax%2B64. So we have x%5E2-16%2Ax%2B-10=%28x-8%29%5E2-74. So completing the square gives x%5E2-16%2Ax%2B-10=highlight%28%28x-8%29%5E2-74%29. Adding 74 from both sides, we get %28x-8%29%5E2=74. Taking the square root of both sides gives system%28x-8=sqrt%2874%29=8.60232526704263%2Cx-8=-sqrt%2874%29=-8.60232526704263%29. system%28x=16.6023252670426%2Cx=-0.602325267042627%29 So the solutions are x=16.6023252670426 and x=-0.602325267042627.
We pick the positive solution, x=8%2Bsqrt%2874%29. So the length is 8%2Bsqrt%2874%29. The width is %288%2Bsqrt%2874%29%29-16=sqrt%2874%29-8.

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
L, length
L-16, width
10, area in square feet

highlight_green%28L%28L-16%29=10%29

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Solving that you find L=8%2Bsqrt%2874%29.