SOLUTION: The length of a rectangle is 4 meters longer than the width. If the area is 29 square meters, find the rectangles dimensions. Round to the nearest tenth of a meter. I know you a

Algebra ->  Radicals -> SOLUTION: The length of a rectangle is 4 meters longer than the width. If the area is 29 square meters, find the rectangles dimensions. Round to the nearest tenth of a meter. I know you a      Log On


   

Question 628329: The length of a rectangle is 4 meters longer than the width. If the area is 29 square meters, find the rectangles dimensions. Round to the nearest tenth of a meter.
I know you are only going to send me the answer, but how you got the answer would help me to understand how to solve a problem like this if I see it again. Thanks.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 4 meters longer than the width.
If the area is 29 square meters, find the rectangles dimensions.
Round to the nearest tenth of a meter.
:
Let W = the width
It says,"The length is 4 meters longer than the width", therefore
L = W+4
:
The area = 29 sq/meters, therefore;
L * W = 29
Replace L with (W+4)
(W+4)*W = 29
W^2 + 4W = 29
A quadratic equation
W^2 + 4W - 29 = 0
Use the quadratic formula to find W
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this problem: x=W, a=1, b=4, c=-29
W+=+%28-4+%2B-+sqrt%284%5E2-4%2A1%2A-29+%29%29%2F%282%2A1%29+
:
W+=+%28-4+%2B-+sqrt%2816-%28-116%29+%29%29%2F2+
:
W+=+%28-4+%2B-+sqrt%28132+%29%29%2F2+
Two solutions, we only want the positive solution
W+=+%28-4+%2B+11.489%29%2F%282%2A1%29+
W = 7.489%2F2
W ~ 3.7 meters is the width
then
3.7 + 4 ~ 7.7 is the length
:
:
Check this by finding the area:
7.7 * 3.7 = 28.49, not 29 because rounded to nearest 10th