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aloeverahe · Feb 18, 2009, 05:56 AM

:confused:I am having trouble with an algebraic problem which I have tried doing for a long time, now it is very close to the due date and I still don't have the answer... :(
The question is:
If x^2 + y^2 + z^2 + 10x + 20y - 30z + 350 = 0, find the value of (x-y-z)^5
If anyone is smart enough to do it, I will thank you a lot.
Very desperate.
Thank you

galactus · Feb 18, 2009, 06:04 AM

Complete the square.

$(x^{2}+10x+25)+(y^{2}+20y+100)+(z^{2}-30z+225)=-350+25+100+125=0$

Factor:

$(x+5)^{2}+(y+10)^{2}+(z-15)^{2}=0$

Now, the values of x,y,z are right in front of you. Find those, then plug them into

$(x-y-z)^{5}$ and you're done.

aloeverahe · Feb 18, 2009, 07:10 AM

Originally Posted by galactus

Complete the square.

Factor:

$(x+5)^{2}+(y+10)^{2}+(z-15)^{2}=0$

Now, the values of x,y,z are right in front of you. Find those, then plug them into

$(x-y-z)^{5}$ and you're done.

What i don't get is how $(x+5)^{2}+(y+10)^{2}+(z-15)^{2}=0$ can change to $(x-y-z)^{5}$ .
I've done the first few steps and i get how you get to the factor but I'm a little confused about how the equations with $^{2}$ can turn into $^{5}$ .
Thanks for you help, I have appreciated it a lot :) :p

galactus · Feb 18, 2009, 07:16 AM

It does not change to $(x-y-z)^{5}$ .

Simply find the values in the equation I showed after completing the square and factoring, then plug them into that.

What makes $(x+5)^{2}=0$ ?

What makes $(y+10)^{2}=0$ ?

What makes $(z-15)^{2}=0$ ?

Easy enough. Find those and plug them in.

I think you're just making this a little harder than need be. See what I mean now?

rwinterton · Feb 18, 2009, 10:47 AM

Any number squared will be greater than zero (unless you are including complex numbers).

So, if (X+5)^2 + (Y+10)^2 + (Z-15)^2 = 0

then

each of the terms must also equal zero (X+5)^2 = 0. There is no other way to add three non-negative numbers and come up with zero.

Solve those three equations for X, Y, and Z and you'll have your answer.