SOLUTION: The area of a rectangle is 60 in ^2. The length is 4 inches greater than the width. Find the length. a) 6 in. b) 3 in. c) 4 in. d) 10 in.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The area of a rectangle is 60 in ^2. The length is 4 inches greater than the width. Find the length. a) 6 in. b) 3 in. c) 4 in. d) 10 in.       Log On


   

Question 310986: The area of a rectangle is 60 in ^2. The length is 4 inches greater than the width. Find the length.

a) 6 in.
b) 3 in.
c) 4 in.
d) 10 in.
Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
lw=60
l=4+w
w^2+4w=60
w^2+4w-60
w=6 l=10
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aw%5E2%2Bbw%2Bc=0 (in our case 1w%5E2%2B4w%2B-60+=+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=%284%29%5E2-4%2A1%2A-60=256.

Discriminant d=256 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+256+%29%29%2F2%5Ca.

w%5B1%5D+=+%28-%284%29%2Bsqrt%28+256+%29%29%2F2%5C1+=+6
w%5B2%5D+=+%28-%284%29-sqrt%28+256+%29%29%2F2%5C1+=+-10

Quadratic expression 1w%5E2%2B4w%2B-60 can be factored:
1w%5E2%2B4w%2B-60+=+1%28w-6%29%2A%28w--10%29
Again, the answer is: 6, -10. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B-60+%29