SOLUTION: I have a problem that seems simple but I'm having an issue solving it. The problem is: A=l*w Length of rectangle 5" more then width. Area is 66". What is length and width? L=w+5

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Question 997800: I have a problem that seems simple but I'm having an issue solving it. The problem is:
A=l*w
Length of rectangle 5" more then width. Area is 66". What is length and width? L=w+5
66= w(w+5) I write this out as w^2+5w-66. Then I factor this so (w+11)(w-6) but I don't know how to finish. Can you help me?
Found 2 solutions by josgarithmetic, MathLover1:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The quadratic should have come as a step in w%5E2%2B5w-66=0 and then factored,
%28w%2B11%29%28w-6%29=0
NOW use Zero Product Rule, and one solution will be meaningless and the other will be meaningful.

Either w+11=0 OR w-6=0.
Which one is needed?

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

A=l%2Aw
Length of rectangle 5 more then width. Area is 66. What is length and width?
L=w%2B5
66=+w%28w%2B5%29
66=w%5E2%2B5w
0=w%5E2%2B5w-66
%28w%2B11%29%28w-6%29=0
solutions:
if %28w%2B11%29=0=>w=-11=> disregard negative solution because we are looking for the width and it cannot be negative number
if %28w-6%29=0=> highlight%28w=6%29
now find the length: L=w%2B5=>L=6%2B5=>highlight%28L=11%29