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If x^2 + y^2 = 5/2 xy, and (x-y)/(x+y)= k where k>0. Expand (x=y)^2. Hence find the value of k.

2006-07-01 21:04:31 · 4 answers · asked by rain-shadow 2 in Education & Reference Homework Help

answer with explaination if possible

2006-07-01 21:13:30 · update #1

4 answers

x^2 + y^2 = 5/2 xy
x²-5xy/2+y²=0
x²-5xy/2+25y²/16=9y²/16
(x-5y/4)²=9y²/16
x-5y/4=±3y/4
x=2y or x=y/2

k=(x-y)/(x+y)
k=(2y-y)/(2y+y) or k=(y/2-y)/(y/2+y)
k=(y)/(3y) or k=(-y/2)/(3y/2)
k=1/3 or k=-1/3
k>0
therefore, k=1/3

2006-07-01 21:16:29 · answer #1 · answered by Pascal 7 · 0 0

lol ... good old math :)

i don't want to make all your homework but i can give you some hints :)
A. use (x-y)(x+y) = x^2 - y^2
B. make substitusions and you must come to this (i made it for x but if you like you can go for y)

4x^2-5X-k=0

more familiar now huh ?

from here (delta) you can find the 2 solutions and keep only the positive one (knowing that k>0)

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He he > king you are stuck because you need to go for a second degree ecuation with X or Y and k not one with X and Y without k.

2006-07-02 04:18:40 · answer #2 · answered by elven_force 2 · 0 0

(X-Y)(X+Y)= X^2-Y^2 SO..K=5/2 XY . EXPAND (X=Y)^2....OK!
( X = Y ) ^ 2.
CAN I HAVE A CRACKER NOW ?

2006-07-02 04:17:03 · answer #3 · answered by DUSTY FOR KING 5 · 0 0

O

2006-07-02 04:09:30 · answer #4 · answered by Rim 6 · 0 0

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