Question 216320
The length of a rectangle is 6 feet more than three times the width. The area of the rectangle is 240 square feet. What are the dimensions? 


Step 1.  Let w be the width and 3w+6 is the length since rectangle is 6 feet more than three times the width.


Step 2.  Area = width * length = w(3w+6) = 240 square feet


Step 3.  Then we can put the equation in Step 2 in quadratic form


{{{3w^2+6w= 240}}}


Subtract 240 to both sides of the equation


{{{3w^2+6w-240=0}}}


Step 4.  We can use the quadratic formula given as


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


where a=3, b=6, and c=-240


*[invoke quadratic "w", 3, 6, -240 ]


So select the positive solution for positive dimensions:


{{{w=8 }}} and {{{3w+6=3*8+6=30}}}


Verify area.... {{{8*30=240}}}...which is a true statement.


Step 5.  Dimensions of the rectangle are a width of 8 feet and a length of 30 feet.


I hope the above steps were helpful.


For FREE Step-By-Step videos in Introduction to Algebra, please visit 

http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit 

http://www.FreedomUniversity.TV/courses/Trigonometry.


Good luck in your studies!


Respectfully,
Dr J