SOLUTION: the area of a rectangle is 102 square inches. the length of the rectangle is 5 inches greater than twice the width. find the dimensions of the rectangle

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Question 385281: the area of a rectangle is 102 square inches. the length of the rectangle is 5 inches greater than twice the width. find the dimensions of the rectangle
Found 2 solutions by mananth, richard1234:
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
the area of a rectangle is 102 square inches. the length of the rectangle is 5 inches greater than twice the width. find the dimensions of the rectangle
...
width = w in
length = 2w+5 in
..
L*W=Area
w(2w+5)=102
2w^2+5w=102
-102 -102
2w^2+5w-102=0
2w^2+17w-12w-102=0
w(2w+17)-6(2w+17)=0
(w-6)(2w+17)=0
w= 6 inches the width
Length=2x+5=12+5=17 inches
....
m.ananth@hotmail.ca

Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website!
It's possible to guess and check by taking the factors of 102 and seeing that (6, 17) works.

Or, we can convert this into an elementary algebra problem by assuming the width is x and the length is 2x + 5. Then,
x(2x + 5) = 102
2x^2 + 5x - 102 = 0
(2x + 17)(x - 6) = 0
x = -17/2 or x = 6 --> x = 6 because we take the positive value.