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.
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.
why is the length x+1? shouldn't it be width=1/2L+1?
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
cool, thanks. I know that's the right answer, didn't know how to get there.
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?
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 ?
I know that, I mean the 1 in the original 1/2L + 1.
Okay, forget it, looked at it again now that I'm awake and get it.
To find the dimensions of the rectangle, we need to set up equations based on the given information.
Let's start by assigning variables to the length and width of the rectangle. Let's call the length "L" and the width "W".
We know that the area of a rectangle is equal to the length multiplied by the width:
Area = Length * Width
Given that the area is 420 sq. meters, we can write the equation as:
420 = L * W ---(equation 1)
We are also given that the width is 1 meter longer than half the length. Mathematically, we can represent it as:
W = (1/2)*L + 1 ---(equation 2)
Now, we can solve this system of equations to find the values of L and W.
Substitute equation 2 into equation 1:
420 = L * [(1/2)*L + 1]
Now, simplify the equation:
420 = (1/2)*L^2 + L
Multiply both sides of the equation by 2 to eliminate the fraction:
840 = L^2 + 2L
Rearrange the equation in standard quadratic form:
L^2 + 2L - 840 = 0
Factor this quadratic equation:
(L + 30)(L - 28) = 0
Setting each factor equal to zero, we get:
L + 30 = 0 or L - 28 = 0
Solving both equations, we find:
L = -30 or L = 28
Since the length cannot be negative, we discard L = -30. Therefore, the length of the rectangle is L = 28 meters.
Substitute this value back into equation 2 to find the width:
W = (1/2)*28 + 1
W = 14 + 1
W = 15 meters
So, the dimensions of the rectangle are length = 28 meters and width = 15 meters.