SOLUTION: The width of a rectangle is 3 ft less than the length. The area of the rectangle is 88 ft2. Find the length L and width W of the rectangle

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Question 1179368: The width of a rectangle is 3 ft less than the length. The area of the rectangle is 88 ft2. Find the length L and width W of the rectangle
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

if the width of a rectangle is 3ft less than the length, then we have
W=L-3ft....eq.1
if the area of the rectangle is 88ft%5E2, we have
L%2AW=88........substitute W
L%2A%28L-3%29=88
L%5E2-3L=88
L%5E2-3L-88=0...factor
L%5E2-11L%2B8L-88=0
%28L%5E2-11L%29%2B%288L-88%29=0
L%28L-11%29%2B8%28L-11%29=0
%28L+-+11%29+%28L+%2B+8%29+=+0
=>L=11 or L=-8-> disregard negative solution for length

go to
W=L-3....eq.1, substitute L=11
W=11-3
W=8


the length of the rectangle L=11ft and width of the rectangleW=8ft


Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


Mentally, in five seconds or less: 88 = 11*8

Algebraically....

x = length
x-3 = width

The area (length times width) is 88:

x%28x-3%29=88
x%5E2-3x=88
x%5E2-3x-88=0
Factor the quadratic:
%28x-11%29%28x%2B8%29=0

The length is either 11 or -8; since -8 makes no sense, it is 11. And then the width is 8.

Note that, in the formal algebraic solution, it was necessary to factor a quadratic equation by finding two numbers whose difference is 3 and whose product is 88; and that is what was required to solve the original problem mentally.

So the formal algebraic solution didn't make solving the problem any easier....

Nevertheless, it is a good exercise for a beginning algebra student to be able to set the problem up and solve it using formal mathematics.