SOLUTION: The length of a rectangle is 7 centimeters less than four times its width. Its area is 15 square centimeters. Find the dimensions of the retangle. Width = Length=

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: The length of a rectangle is 7 centimeters less than four times its width. Its area is 15 square centimeters. Find the dimensions of the retangle. Width = Length=      Log On


   

Question 344479: The length of a rectangle is 7 centimeters less than four times its width. Its area is 15 square centimeters. Find the dimensions of the retangle.
Width =
Length=
Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
a=lw
a=15
l=4w-7
15=4w^2-7w
0=4w^2-7w-15
w=3 since it cant be negative
l=5
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aw%5E2%2Bbw%2Bc=0 (in our case 4w%5E2%2B-7w%2B-15+=+0) has the following solutons:

w%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-7%29%5E2-4%2A4%2A-15=289.

Discriminant d=289 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--7%2B-sqrt%28+289+%29%29%2F2%5Ca.

w%5B1%5D+=+%28-%28-7%29%2Bsqrt%28+289+%29%29%2F2%5C4+=+3
w%5B2%5D+=+%28-%28-7%29-sqrt%28+289+%29%29%2F2%5C4+=+-1.25

Quadratic expression 4w%5E2%2B-7w%2B-15 can be factored:
4w%5E2%2B-7w%2B-15+=+4%28w-3%29%2A%28w--1.25%29
Again, the answer is: 3, -1.25. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B-7%2Ax%2B-15+%29