SOLUTION: The length of a rectangle is 9 inches more than the width. Its area is 112 square inches. Find the dimensions of the rectangle.

Question 227095: The length of a rectangle is 9 inches more than the width. Its area is 112 square inches. Find the dimensions of the rectangle.
Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 9 inches more than the width. Its area is 112 square inches. Find the dimensions of the rectangle.

Step 1. The area A of a rectangle is given as A=length*width.

Step 2. Let w be the width.

Step 3. Let w+9 be the length

Step 4. Then, A=w(w+9)=112 or w^2+9w-112=0

Step 5. To solve, use the quadratic formula given as

where a=1, b=9, and c=-112

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=529 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 7, -16. Here's your graph:

For w=7 , then w+9=16 and A=7*16=112 which is a true statement.

Step 6. ANSWER: The width is 7 inches and the length is 9 inches.

I hope the above steps were helpful.

For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J

drjctu@gmail.com

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