Asked by Bentley

A rectangle has a length that is 1 less than twice its width. It's area is 21. Find its perimeter.
Answered by Reiny
width --- x
length ---2x - 1
x(2x-1) = 21
2x^2 -x -21 = 0
((x+3)(2x-7)=0
x = -3 , silly or
x = 7/2 which is 3.5

width = 3.5
length = 6

perimeter = 2(3.5+6) = 19
Answered by Kuai

A = lw

l = 2x-1

w = x

21 = x(2x -1)

21 = 2x^2 -x

2x^2-x -21 =0

(x +3)(2x -7)

x = 7/2 = 3.5

2(7/2)-1 = 6

l = 6; w = 7/2

p = 2l + 2w

p = 2(6) + 2(3.5) = 19
Answered by Anonymous
How did you get the value of x from (x+3)(2x-7)=0
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