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Old 07-26-2010   #3
MAS1

 
Join Date: Dec 2008
Posts: 249
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Quote:
Originally Posted by twisted View Post
I need some help figuring out how to find the dimensions to this question:

A rectangle has perimeter 64cm and area 23cm^2. Solve the following system of equations to find the rectangle's dimensions.

a) l = 23/w
l + w = 32

b) solve the system of equations
x^2 + y^2 = 1
xy = 0.5

Please and thank you! I know the answer, I just don't know how to do the question.
Since Mr Hui answered part a), I'll try part b).

Since xy = 0.5 then x = 1/(2y)
Substituting for x gives:
(1/(2y))^2 + y^2 = 1
1/(4y^2) + y^2 = 1
(1 + 4y^4)/(4y^2) = 1
1 + 4y^4 = 4y^2
4y^4 - 4y^2 + 1 = 0
y^4 - y^2 + 1/4 = 0
(y^2 - 1/2)(y^2 - 1/2) = 0
y^2 - 1/2 = 0
y^2 = 1/2
y = + or - sqrt(1/2)
x = + or - sqrt(1/2)
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