Author Topic: Some tough crunching (Read 744 times) Tweet Share

crabby · « **on:** February 17, 2011, 07:36:02 pm »

Need help with this maths question.

If $x^2+y^2+z^2+10x+20y-30z+350=0$ , what is the value of $(x-y-z)^5$ ?

evaever · « **Reply #1 on:** February 18, 2011, 11:04:25 pm »

Quote from: crabby on February 17, 2011, 07:36:02 pm

Need help with this maths question.

If $x^2+y^2+z^2+10x+20y-30z+350=0$ , what is the value of $(x-y-z)^5$ ?

$(x+5)^2+(y+10)^2+(z-15)^2=0$
$x=-5,y=-10$ and $z=15$
.: $(x-y-z)^5=-10^5$

shinny · « **Reply #2 on:** February 18, 2011, 11:11:52 pm »

Quote from: evaever on February 18, 2011, 11:04:25 pm

Quote from: crabby on February 17, 2011, 07:36:02 pm
Need help with this maths question.

If $x^2+y^2+z^2+10x+20y-30z+350=0$ , what is the value of $(x-y-z)^5$ ?
$(x+5)^2+(y+10)^2+(z-15)^2=0$
$x=-5,y=-10$ and $z=15$
.: $(x-y-z)^5=-10^5$

In an exam, I'd probably add an explanation in between the first and second step as to why they each bracket must equal zero. Simply put for whoever didn't understand it, the square of each bracket has to be a positive number. Note that we're adding together each bracket though. The only way you can get a total of zero from adding three positive numbers is if each bracket is equal to zero, so then you'd solve ${x+5}^{2}=0$ and so on individually in it's own line in proper exam working out. Generally it's best to play safe and explain every step rather than jumping straight to giving the answer in VCE, even if it seems quite obvious to you.

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Author Topic: Some tough crunching (Read 744 times) Tweet Share

crabby

Some tough crunching

evaever

Re: Some tough crunching

shinny

Re: Some tough crunching

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