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Old 03-07-06, 08:05 PM   #1
Snuffleupagus
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Dumb Math Question (no...really...it's dumb)

So yeah, I didn't pay attention in school, and now I'm self teaching. My math book gives me this equation:

(y/2+z/3) * (y/2+z/3)

My logic is as follows:

y/2*y/2=y^2/4

y/2*z/3=yz/6

z/3*z/3= z^2/9

So my mid point is y^2/4 + yz/6 + yz/6 + z^2/9

Which simplifies to y^2/4 + yz/3 + z^2/9

My book OTOH tells me that the equation simplifies to y^2/4 + y^z/3 + z^2/9

Plugging in numbers for the variables leads me to...frustration I just need a brief explanation as to why the problem simplifies as it does.

Danke.
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Old 03-07-06, 08:09 PM   #2
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Your book evidently has a typo.
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Old 03-07-06, 08:15 PM   #3
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So my answer of y^2/4 + yz/3 + z^2/9 is correct given a problem of (y/2+z/3) * (y/2+z/3) I take it?
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Old 03-07-06, 08:15 PM   #4
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Hmmm I got the same result you did. Are you sure it could not be a typo?
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Old 03-07-06, 08:21 PM   #5
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Snuffleupagus, your answer is correct, and more importantly, your reasoning is sound.
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Old 03-07-06, 08:23 PM   #6
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Your answer looks right to me. You wouldn't raise y to the z power in that instance.
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Old 03-07-06, 08:26 PM   #7
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Thanks for the help!

Now back to work...
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Old 03-08-06, 06:59 AM   #8
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Wrong answers in the book can be very frustrating.

Be sure you point it out to your professor/teacher. I had one professor that gave extra points if you caught an error in the book!

You will get to the point that you are so confident in your answers, if the book has a different answer than you do, you will always believe the BOOK is wrong.

Then you will get your degree.

Now for some math fun. You are correct,

(a+b)^2 = a^2 + 2ab + b^2

Are you familiar with Pascal's triangle? The sum of the two numbers on the row above equals the number between those two numbers below. It relates to the problem you solved. It looks like this:

____1
__1 2 1
_1 3 3 1
1 4 6 4 1
etc...

As we found before: (a+b)^2 = a^2 + 2ab + b^2. Note the coefficients of each term are
1,2,1. The same as the second row in the triangle. Now watch this:

(a+b)^3 = a^2 + 3(a^2)b + 3 a(b^2)+b^3. The coefficients of each term are now
1,3,3,1. The same as the third row in the triangle!

And so on, for any power of any binomial.

This may save you some time during a test some day.
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Old 03-08-06, 08:55 AM   #9
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Quote:
Originally Posted by eubi
(a+b)^3 = a^3 + 3(a^2)b + 3 a(b^2)+b^3. The coefficients of each term are now
1,3,3,1. The same as the third row in the triangle!
Fixed
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Old 03-08-06, 10:09 AM   #10
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Quote:
Originally Posted by eubi
You will get to the point that you are so confident in your answers, if the book has a different answer than you do, you will always believe the BOOK is wrong.
...after verifying that the book's answer isn't (a) an equivalent answer expressed in a different form or (b) an alternative answer when asked to only look for one answer but when multiple answers exist.

You will also develop a sense for "reasonableness" of answers. In your original post, the book's answer was clearly wrong. The thing that took any time at all to determine was whether your answer was correct.

Quote:
Originally Posted by eubi
Are you familiar with Pascal's triangle? The sum of the two numbers on the row above equals the number between those two numbers below. It relates to the problem you solved. It looks like this:

____1
___1 1
__1 2 1
_1 3 3 1
1 4 6 4 1
etc...
Pascal's triangle (slight correction above) is useful for a number of things. Including what eubi pointed out. Good stuff.
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Old 03-08-06, 10:27 AM   #11
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Perhaps i'm missing something here (and most likely i am), but isn't the most simplified version of this:

(yz/6)^2

maybe i just don't understand exactly what they're looking for. or maybe my math is wrong. see what happens when you pick your college because they sent you a letter saying your SAT scores were high enough to get in? curse you UNO, curse you!
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Old 03-08-06, 10:32 AM   #12
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Quote:
Originally Posted by Ubie
Perhaps i'm missing something here (and most likely i am), but isn't the most simplified version of this:

(yz/6)^2

maybe i just don't understand exactly what they're looking for. or maybe my math is wrong. see what happens when you pick your college because they sent you a letter saying your SAT scores were high enough to get in? curse you UNO, curse you!
No, you don't multiply the two expressions again, you add them. Hence it would be yz/6+yz/6 = 2yz/6 which simplifies to yz/3. Go find an 11th grader and (s)he'll explain it better.

On paper it'd look something like this:

(y/2 + z/3)*(y/2 + z/3)

y^2/4
blank blankyz/6
blank blankyz/6
blank blank blank blankz^2/9
+__________________
y^2/4 + 2yz/6 + z^2/9

simplified: y^2/4 + yz/3 + z^2/9
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Last edited by jyossarian; 03-08-06 at 10:38 AM.
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Old 03-08-06, 10:37 AM   #13
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Quote:
Originally Posted by Ubie
Perhaps i'm missing something here (and most likely i am), but isn't the most simplified version of this:

(yz/6)^2

maybe i just don't understand exactly what they're looking for. or maybe my math is wrong. see what happens when you pick your college because they sent you a letter saying your SAT scores were high enough to get in? curse you UNO, curse you!
I think you looked at the original equation and accidently put in multiplication signs where the plus signs are.
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Old 03-08-06, 10:57 AM   #14
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Quote:
Originally Posted by jyossarian
No, you don't multiply the two expressions again, you add them. Hence it would be yz/6+yz/6 = 2yz/6 which simplifies to yz/3. Go find an 11th grader and (s)he'll explain it better.

On paper it'd look something like this:

(y/2 + z/3)*(y/2 + z/3)

y^2/4
blank blankyz/6
blank blankyz/6
blank blank blank blankz^2/9
+__________________
y^2/4 + 2yz/6 + z^2/9

simplified: y^2/4 + yz/3 + z^2/9
d'oh! you're right!

***edit, everything i had entered is wrong, trying again below***

so, the most simplified, straightforward thing i can come up with is:

((3y+2z)/6)^2

Last edited by Ubie; 03-08-06 at 11:03 AM.
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Old 03-08-06, 06:21 PM   #15
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Good reading.

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Old 03-09-06, 07:01 AM   #16
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Quote:
Originally Posted by eubi
(a+b)^3 = a^3 + 3(a^2)b + 3 a(b^2)+b^3.
Argh!

What I get for trying to teach math during break!

Good catch jyossarian!
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