SOLUTION: What is the area of a rectangle that is 5 inches longer than it is wide and if you double the length and subtract 2 inches from the width the area is increased by 52 inches squared

Algebra ->  Rectangles -> SOLUTION: What is the area of a rectangle that is 5 inches longer than it is wide and if you double the length and subtract 2 inches from the width the area is increased by 52 inches squared      Log On


   

Question 527520: What is the area of a rectangle that is 5 inches longer than it is wide and if you double the length and subtract 2 inches from the width the area is increased by 52 inches squared?
I tried...
(x-2)(2x+10)=x^2+5x+52
but I don't thin Im setting it up right...
Can you help me set it up?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You are properly setting it up.
The original rectangle has a width that you called x
and we should add a requirement that it should be a real number, and that
x%3E0
The length of the original rectangle is x%2B5
Its area is %28x%28x%2B5%29=x%5E2%2B5x
The new rectangle has a length of 2%28x%2B5%29=2x%2B10
a width of x-2
(so we are also going to ask for x%3E2)
The area of the new rectangle is
%28x-2%29%282x%2B10%29=2x%5E2%2B6x-20
Your equation transforms into
2x%5E2%2B6x-20=x%5E2%2B5x%2B52 and into
(((x^2+x-72=0}}}
which can be solved by factoring, or by completing the square, or by using the quadratic formula.
One of the solutions for x is negative. The other solution is the width of the original rectangle, which you would need to use to find the area of the original rectangle.