math question

[Glowwyrm]

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Can you look at this and tell me if it makes any sense to you? I'm going to use x^2 for x squared.

If y = x^2, z = x^2, and w = xy, then y^2 + z^2 + w^2 =

And their solution:

w = ys = x(x^2) = x^2 (I'm assuming by ys they actually mean yx but since when does x times x squared equal x squared?)

y^2 + z^2 + w^2 = (x^2)^2 + (x^3)^2 + (x^3)^2

= x^4 + x^6 + x^6 = x^4 + 2x^6

What do y'all get when you do this problem? I keep getting 2x^4 + x^6 instead of x^4 +2x^6
Thanks.

 

[acetyl]

I get the same thing as you, 2x^4 + x^6.

This is wrong:
w = ys = x(x^2) = x^2 (I'm assuming by ys they actually mean yx but since when does x times x squared equal x squared?)

x(x^2)= x^3

 

[phospho]

Can you look at this and tell me if it makes any sense to you? I'm going to use x^2 for x squared.

If y = x^2, z = x^2, and w = xy, then y^2 + z^2 + w^2 =

And their solution:

w = ys = x(x^2) = x^2 (I'm assuming by ys they actually mean yx but since when does x times x squared equal x squared?)

y^2 + z^2 + w^2 = (x^2)^2 + (x^3)^2 + (x^3)^2

= x^4 + x^6 + x^6 = x^4 + 2x^6

What do y'all get when you do this problem? I keep getting 2x^4 + x^6 instead of x^4 +2x^6
Thanks.

loooool!
I've been working on this dam thing for the last 15 minutes and I'm getting the same as you! It isn't working...I've tried every single thing I can possibly think about...wow!

For what it's worth though, their answer seems reasonably correct (I hope at least) because:

lets say x=1 hypothetically speaking
then y will equal 1
z will equal 1
w will equal 1
that would mean that (again, assuming x=1):
y^2 + z^2 + w^2 = 3
x^4 + x^4 + (xy)^2 = 3
2x^4 + x^6 = 3

now their answer x^4 + 2x^6 ALSO equals 3! I have no idea, i feel really stupid right now...lol

 

[Glowwyrm]

Ok good. At least I know I'm not crazy.

 

[Lnguyen1231]

PLEASE NOTE: x^m*x^n=x^m+n; a^r*b^r=(ab)^r

 

[Lnguyen1231]

tried soo many times and still got the same answer

 

[Lnguyen1231]

look at this u guys: y^2 + z^2 + w^2 IS NOT EQUAL TO (x^2)^2 + (x^3)^2 + (x^3)^2

BUT IT SHOULD BE: y^2+z^2+w^2=(x^2)^2+(x^2)^2+(x*x^2)^2=x^4+x^4+x^6=2x^4+x^6 (this makes more sense....but let's try it again to see if we get the same answer as in the book...i got the same the first time but then i tried to write it out....i lost track of it...argggg)0

 

[MahSpoon]

I was just browsing this area of the forum for fun, and after taking the MCAT, i'm glad i don't have to fiddle with this stuff...are they seriously asking this on the PCAT? pretty useless and annoying algebra o_P

 

[Lnguyen1231]

this problem is very simple BUT..THE ANSWER drives everyone crazy...making us all feel stupid when they make some kind of mistakes like this one...Calculus in the pcat is just for school entry and after that you'll never have to see it or use it. they just want to make it tough just to get into pharm school, that's all. When u work in the pharmacy, the math IS VERY SIMPLE AND BASIC...Very easy math.

 

[cdpiano27]

The book solution is just plain wrong

It is very easy, and Algebra I level!!!

If y = x^2, z = x^2 and w = xy

this implies that y^2 = x^4, z^2 = x^4 and w^2 = x^2 * y^2 = x^2 * x^4 = x^6 \.

Hence y^2 + z^2 + w^2 = x^4 + x^4 + x^6 = x^6 + 2x^4
factored out this equals x^4 * (2 + x^2).

 

[Lnguyen1231]

Hi cdpiano27!
I think everyone knows how to solve this problem pretty well...HOWEVER, THE answer in the book is what drives everyone going crazyyyyyyyyyyy....after all, the problem is just plain simple...THEY BETTER STOP MAKING MISTAKES LIKE THAT....

 

[Glowwyrm]

Yeah. I don't think anyone that's even had high school math will see this as a difficult problem. I just wanted to make sure the book was wrong and it wasn't me making some mistake that I just wasn't catching.