Question about one specific Math question!! June 06

Test Preparation SAT Preparation
21

<p>i'm sure they would have specified x, y, z cannot equal 0
bcuz even if these integers were negative.
for example."" 2^2+-2^2+1^2=9 so the answer would be 1 . "" this is not true. bcuz it would be (2)^2 + (-2)^2 + (1)^2 = 5
everyone knows that a negative number raised to an even integer is positive. so i'm sure they specified x y and z do not equal 0.</p>

22

<p>yes, as I said, I'm fairly certain it was positive intergers, and therefore the solution was indeed 5.</p>

23

<p>ah see,</p>

<p>(x+y+z)^2= (x^2 + y^2 + z^2) + 2(xy + xz + yz)
I II III</p>

<p>since II=9</p>

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<p>If x,y,,z are positive integers then I>9 and hence x+y+z >3</p>

<p>had x,y,z been negative,then the value of III would have oscillated between <or> zero.... and hence there would have been endless possibilities</or></p>

<p>therefore the question must have stated the limit for x,y,z to be positive integers
ididnt gave the SAT though, yet i blv 5 shud be the answer</p>

24

<p>sorry i 4got to mention above I II III</p>

<p>I=(x+y+z)^2
II= (x^2 + y^2 + z^2)
III= 2(xy + xz + yz)</p>

25

<p>there was no need of that. you just made it overly complicated for half the people on this forum.</p>

26

<p>i think the question restricted it to be a positive integer so zero and negative numbers are out of the picture</p>

27

<p>i didnt have this section, i dont think. Did you all have the passage about a black businessman, and the one about the weathermen?</p>

28

<p>no i did not, where you in the east or west coast? (I'm east)</p>

29

<p>are you** (10 chac)</p>