Question

If x and Y are the roots of equation x^2+4x-4=0 , 1/(x-2)^3 + 1/(y-2)^3=?

Answer 1

can we put z instead of x??

Answer 2

ok

Answer 3

(x-z)(x-y) = x^2+4x-4
zy = 16/2 = -4
z+y = -4

Answer 4

-5/8

Answer 5

\[ \frac{(y-2)^3 +(z-2)^3}{((y-2)(z-2))^3} =
\frac{(y+z-4)((y-2)^2 - (y-2)(z-2) + (z-2)^2 }{(yz - 2 (y+z) + 4)^3}\]

Answer 6

X=2 sqrt(2)-2
Y=-2 sqrt(2)-2
REPLACE THEM THERE U GET 1/((-2)sqrt(2)-2-2)^3+1/(2sqrt(2)-2-2)^3=-0.625

Answer 7

I think we need to evaluate y^2+z^2

\[ \left ( \frac{-b + \sqrt{b^2 - 4ac}}{2a} \right )^2 + \left ( \frac{-b - \sqrt{b^2 - 4ac}}{2a} \right )^2 = \frac{2b^2 + 2(b^2 - 4ac)}{4a^2} \\
\\ = \frac{4b^2 - 8ac}{4a^2} = \frac{4.4^2 - 8.1.4}{4} \]

evaluate rest ... and put values ... and there might me some errors ... please check yourself

Answer 8

@experimentX umissunderstood the question i think !
open study did u found the roots ! like those
X=2 sqrt(2)-2
Y=-2 sqrt(2)-2
!?

Answer 9

well ... i believe this is more general way of solving problem
unless you are still junior at secondary school

Answer 10

no!

Answer 11

yeh sorry any way
we trying to solve this question now right ^_^

Answer 12

x+y=-4 xy=-4

Answer 13

keep trying ... :)

Answer 14

@experimentX is that the first step

Answer 15

yes!!
you can find it by quadratic formula ... sum of roots and product of roots!!
but verify!! I'm a bit inaccurate in calculations!!

Answer 16

thanzz a lot @Callisto @experimentX @badi

Answer 17

Don't just copy the answer!!!! Do it on you own!!!!

Answer 18

ok

Answer 19

thanzz for giving me an idea