SOLUTION: The length of a rectangle is five more than twice its width. The area of the rectangle is 52 sqm. Find the dimensions of the rectangle.

Question 358404: The length of a rectangle is five more than twice its
width. The area of the rectangle is 52 sqm. Find the
dimensions of the rectangle.
Found 2 solutions by mananth, sudhanshu_kmr:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let width be x
length = 2x+5
area = 52 m^2
Area = L*W
x(2x+5)=52
2x^2+5x=52
2x^2+5x-52=0
2x^2-8x+13x-52=0
2x(x-4)+13(x-4)=0
(x-4)(2x+13)=0
x=4 x = -6.5(ignore)
width = 4
length = 2x+5 = 13
...
m.ananth@hotmail.ca
Answer by sudhanshu_kmr(1152) (Show Source): You can put this solution on YOUR website!

Let width is x, then length = 2x+5
x(2x+5) = 52
=> 2x^2 +5x -52 = 0
=> 2x^2 +13x - 8x -52 = 0
=> x(2x + 13) -4(2x + 13) = 0
=> (x-4) (2x+13) = 0
hence x =4, 2x+5 = 13

dimensions are 4 and 13.
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