Asked by Randy

The area of a rectangle is 420 sq. meters. The width is 1 meter longer than half the length, find the dimensions of the court. Not sure how to do this, although I know the answer.

Answers

Answered by Reiny
let the width be x m
then the length is x+1
x(x+1) = 420
x^2 + x - 420 = 0
(x+21)(x-20) = 0

take it from here.
Answered by Randy
why is the length x+1? shouldn't it be width=1/2L+1?
Answered by Reiny
You are right, did not read it carefully enough.

let the length be l
then width is l/2 + 1
l(l/2 + 1 ) = 420
l^2/2 + l = 420
l^2 + 2l = 840
l^2 + 2l - 840 = 0
(l+30)(l-28) = -
l = 28 or l= -30, but x has to be clearly positive
length = 28
widht = 15
check :15x28 = 420

Answered by Randy
cool, thanks. I know that's the right answer, didn't know how to get there.
Answered by Randy
Okay, so...
L(1/2L + 1) =420
L^2/2L+1=420
L^2 + 2L = 840
L^2 + 2L -840=0
factor
(L-30)(L+28)=0
what happens to the +1?
Answered by Reiny
well if l= 28
then 1/2 of l is 14
but it said 1 more than 1/2 the length
so what is 14 + 1 ?
Answered by Randy
I know that, I mean the 1 in the original 1/2L + 1.
Answered by Randy
Okay, forget it, looked at it again now that I'm awake and get it.
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