SOLUTION: the width of a rectangle is 5ft less than the length. the area is 6ft^2

Algebra ->  Rectangles -> SOLUTION: the width of a rectangle is 5ft less than the length. the area is 6ft^2       Log On


   

Question 221566: the width of a rectangle is 5ft less than the length. the area is 6ft^2
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
The width of a rectangle is 5 ft less than the length. The area is 6ft^2

I assume you want to find the dimensions of the rectangle.

Step 1. Let L be the length.

Step 2. Let L-5 be the width.

Step 3. Area A=L%28L-5%29=6

Step 4. Solving A in Step 3 yields the following steps.

L%5E2-5L=6

Subtract 6 from both sides to get a quadratic equation

L%5E2-5L-6=6-6

L%5E2-5L-6=0

Step 5. To solve equation in Step 5, use the quadratic formula given below

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a=1, b=-5, and c=-6

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aL%5E2%2BbL%2Bc=0 (in our case 1L%5E2%2B-5L%2B-6+=+0) has the following solutons:

L%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=%28-5%29%5E2-4%2A1%2A-6=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+49+%29%29%2F2%5Ca.

L%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+49+%29%29%2F2%5C1+=+6
L%5B2%5D+=+%28-%28-5%29-sqrt%28+49+%29%29%2F2%5C1+=+-1

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



Using the positive solution L=6, then L-5=1. Area A=6*1=6 which is a true statement.

Step 6. ANSWER: The dimensions of a rectangle are 6 feet and 1 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