SOLUTION: The area of a rectangle is 52 and the length is 5 less than twice the width. What are the dimensions of the rectangle?

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Question 887133: The area of a rectangle is 52 and the length is 5 less than twice the width. What are the dimensions of the rectangle?
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Step 1, try the factors of 52 in pairs.
----
1*52 NG
2*26 NG
etc
==============
If that doesn't work:
L*W = 52
L = 2W - 5
Sub for L in the 1st eqn
(2W - 5)*W = 52

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=441 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 6.5, -4. Here's your graph:

=======================
W = 6.5
L = 8
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