SOLUTION: a rectangle is 4 cm longer than it is wide, and it area is 117 cm squared. find its dimensions.

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Question 224584: a rectangle is 4 cm longer than it is wide, and it area is 117 cm squared. find its dimensions.
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle is 4 cm longer than it is wide, and it area is 117 cm squared. Find its dimensions.

Step 1. Let w be the width

Step 2. Let w+4 be the length

Step 3. Let Area A=w(w+4)=117 where the area is the product of the length and the width.

Step 4. Then w%5E2%2B4w-117=0 where we subtracted 117 from both sides of the equation to get a quadratic equation.

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

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

where a=1, b=4 and c=-117

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B-117+=+0) has the following solutons:

x%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-117=484.

Discriminant d=484 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+484+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+484+%29%29%2F2%5C1+=+9
x%5B2%5D+=+%28-%284%29-sqrt%28+484+%29%29%2F2%5C1+=+-13

Quadratic expression 1x%5E2%2B4x%2B-117 can be factored:
1x%5E2%2B4x%2B-117+=+1%28x-9%29%2A%28x--13%29
Again, the answer is: 9, -13. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B-117+%29



Selecting the positive solution we have w=9 then w+4=13 and A=9*13=117 which is a true statement.

Step 6. ANSWER: The width is 9 cm and the length is 13 cm.

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.

And good luck in your studies!

Respectfully,
Dr J
http://www.FreedomUniversity.TV