SOLUTION: The length of a rectangle is 6 cm more than the width. The area is 11 cm^2. Find the length and width.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The length of a rectangle is 6 cm more than the width. The area is 11 cm^2. Find the length and width.      Log On


   

Question 991021: The length of a rectangle is 6 cm more than the width. The area is 11 cm^2. Find the length and width.
Answer by Timnewman(323) About Me  (Show Source):
You can put this solution on YOUR website!
Hi dear,
let L=lenght,
w=width
Then,from your question,
L=w+6-----(1)
and
L*w=11----(2)
Put 1 in 2,
(w+6)w=11
w²+6w-11=0
Solve the resulting quadratic equation as follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aw%5E2%2Bbw%2Bc=0 (in our case 1w%5E2%2B6w%2B-11+=+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=%286%29%5E2-4%2A1%2A-11=80.

Discriminant d=80 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-6%2B-sqrt%28+80+%29%29%2F2%5Ca.

w%5B1%5D+=+%28-%286%29%2Bsqrt%28+80+%29%29%2F2%5C1+=+1.47213595499958
w%5B2%5D+=+%28-%286%29-sqrt%28+80+%29%29%2F2%5C1+=+-7.47213595499958

Quadratic expression 1w%5E2%2B6w%2B-11 can be factored:
1w%5E2%2B6w%2B-11+=+1%28w-1.47213595499958%29%2A%28w--7.47213595499958%29
Again, the answer is: 1.47213595499958, -7.47213595499958. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B6%2Ax%2B-11+%29

Then w=1.5cm,w=-7.3cm
When w=1.9,
L=1.9+6
L=7.9cm
when w=-7.3,
L=-7.3+6
L=-1.3cm