Help! Fiding dimentions from an equation. - XP Math - Forums

 XP Math - Forums Help! Fiding dimentions from an equation.

 07-16-2010 #1 twisted   Join Date: Jul 2010 Posts: 1 Help! Fiding dimentions from an equation. 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.
 07-21-2010 #2 Mr. Hui     Join Date: Mar 2005 Posts: 10,609 l = 23/w l + w = 32 23/w + w = 32 23 + w^2 = 32w w^2 - 32w + 23 = 0 Use the quadratic formula to solve for w: w = 31.264 and 0.73566. l = 1.35930435 and 0.0319852174. __________________ Do Math and you can do Anything!
07-26-2010   #3
MAS1

Join Date: Dec 2008
Posts: 249

Quote:
 Originally Posted by twisted 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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