SOLUTION: The length of a rectangle is 5 in. more than its width. Its area is 75 sq. in. what is the length and width of the rectangle?

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Question 112309: The length of a rectangle is 5 in. more than its width. Its area is 75 sq. in. what is the length and width of the rectangle?
Found 2 solutions by checkley71, solver91311:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
LENGTH=W+5
AREA=L*W
75=(W+5)W
75=W^2+5W
W^2+5W-75=0
USING THE QUADRATIC EQUATION WE GET:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
W=(-5+-SQRT[5^2-4*1*-75])/2*1
W=(-5+-SQRT25+300])/2
W=(-5+-SQRT325)/2
W=(-5+-18.03)/2
W=-5+18.03
W=13.03/2
W=6.515 SOLUTION.
L=6.515+5
L=11.515 SOLUTION.
PROOF:
6.515*11.515=75
75=75

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
We know that the area of a rectangle is given by A=lw. But the problem tells us that l=w%2B5. The length (l) is (=) 5 in (5) more than (+) its width (w).

Since we know that the area is 75 sq. in., we can now write %28w%2B5%29%2Aw=75

Simplifying and solving:

w%5E2%2B5w=75
w%5E2%2B5w-75=0
This one doesn't factor, so use the quadratic formula
w+=+%28-5+%2B-+sqrt%28+5%5E2-4%28-75%29%29%29%2F%282%29+
w+=+%28-5+%2B-+sqrt%28325%29%29%2F2

w+=+%28-5+%2B-+5%2Asqrt%2813%29%29%2F2

One of the possibilities, namely w+=+%28-5+-+5%2Asqrt%2813%29%29%2F2, yields a negative result which is an absurdity when you are trying to find the length of a rectangle's side, therefore, the only valid answer is w+=+%28-5+%2B+5%2Asqrt%2813%29%29%2F2. So now we have the width.

The length is just 5 inches longer, so l+=+%28%28-5+%2B+5%2Asqrt%2813%29%29%2F2%29%2B5
or in simpler terms
l=%285%2B5%2Asqrt%2813%29%29%2F2

Leave your answers in terms of these simplest form expressions containing the radicals. If the problem had asked you to determine the length and width to some number of decimal places accuracy, THEN you would use a calculator and round appropriately, but you have to presume that exact expressions for the lengths are required because the problem asks "what is the length and width" without an accuracy specification.