SOLUTION: The floor area of a certain semirailer is 306 square feet. If It's length is 2 feet more than 4 times its width, what is its width?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: The floor area of a certain semirailer is 306 square feet. If It's length is 2 feet more than 4 times its width, what is its width?      Log On


   

Question 1115664: The floor area of a certain semirailer is 306 square feet. If It's length is 2 feet more than 4 times its width, what is its width?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

A=L%2AW

given:
the floor area is
L%2AW=306...eq.1
it's length L is 2 feet more than 4 times its width W
L=4W%2B2...eq.2...substitute in eq.1

%284W%2B2%29%2AW=306...solve for W

4W%5E2%2B2W=306
4W%5E2%2B2W-306=0...simplify
2W%5E2%2BW-153=0...factor
2W%5E2%2B18W-17W-153=0
%282W%5E2%2B18W%29-%2817W%2B153%29=0
2W%28W%2B9%29-17%28W%2B9%29=0
%28W+%2B+9%29+%282W+-+17%29+=+0
solutions:
if %28W+%2B+9%29++=+0=>W=-9-> since we need width, disregard negative solution
if %282W+-+17%29+=+0=>W=17-> the width