RE: Any Math Wizards Care to Take A Crack At This Problem? (Full Version)

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LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:31:33 AM)

quote:

ORIGINAL: MissAsylum

Photo of the problem

[image]local://upfiles/739995/D7E83355ECEA42B9835BDAB911F7FFEB.jpg[/image]


I'll try it by sight :

Each term has X * Y^2, right?
That leaves:

X*(Y^2) * {(3X)^.25  +  (3^5*X)^.25  +  [(2^4 * 3) * X]^.25}
X*(Y^2) * {(3X)^.25  +  3*(3X)^.25  +  2*(3 X)^.25}
And 1+3+2 = 6,

So: X * (Y^2) * 6* (3X)^.25 

Final answer.  There are variations, of course.  Put the X back into the quarter root and so on.



MasterG2kTR -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:34:34 AM)

Yup...I was just gonna post that answer for you....SERIOUSLY!! I just saw this and worked it out cuz that just happens to be the math chapter I will be testing for tomorrow.

But yes.....6xy^2 4^√3x this is the same answer I came up with.



VaguelyCurious -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:35:10 AM)

I left everything in the quarter root, and got ( 294 (x^5) (y^8) )^(1/4)



LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:36:26 AM)

quote:

ORIGINAL: MissAsylum

So.... 6xy^2 4^√3x.


I think.

YES!  You GO, girl!  Except the notation on a keyboard really doesn't allow for fractional roots the way you're trying to show.  Either (3x)^.25  OR (3x)^(1/4) -  The 2nd is my prefernce.

Notation is EVERYTHING!



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:40:16 AM)


quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: MissAsylum

So.... 6xy^2 4^√3x.


I think.

YES!  You GO, girl!  Except the notation on a keyboard really doesn't allow for fractional roots the way you're trying to show.  Either (3x)^.25  OR (3x)^(1/4) -  The 2nd is my prefernce.

Notation is EVERYTHING!


Ugh. Please...don't remind me.



GreedyTop -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:41:07 AM)

*sigh* I'm such a bonehead when it comes to math... I wish my Grandfather was around so I could have run this by him...



LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:41:07 AM)

quote:

ORIGINAL: VaguelyCurious

I left everything in the quarter root, and got ( 294 (x^5) (y^8) )^(1/4)

Well, your integer is WAY off.
6 * (6 ^ .25) = (6 ^4)^.25 * (6^.25) = (6^5)^.25 = 7776 ^ .25



LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:43:29 AM)

quote:

ORIGINAL: MissAsylum


quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: MissAsylum

So.... 6xy^2 4^√3x.


I think.

YES!  You GO, girl!  Except the notation on a keyboard really doesn't allow for fractional roots the way you're trying to show.  Either (3x)^.25  OR (3x)^(1/4) -  The 2nd is my prefernce.

Notation is EVERYTHING!


Ugh. Please...don't remind me.

And get those parens in there!

6xy^2 could be just about anything!

inc. (6^2)(x^2)(y^2)



VaguelyCurious -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:52:53 AM)


quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: VaguelyCurious

I left everything in the quarter root, and got ( 294 (x^5) (y^8) )^(1/4)

Well, your integer is WAY off.
6 * (6 ^ .25) = (6 ^4)^.25 * (6^.25) = (6^5)^.25 = 7776 ^ .25


Where are you getting all the sixes from?

I made the first term (3(x^5)(y^8))^(1/4)
the second term (243(x^5)(y^8))^(1/4)
and the third term (48(x^5)(y^8))^1/4

....and then I'm an idiot, because you can't add them straight up. In my defence, I'm very tired...

Whoops.

Sorry. [8D]



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:54:19 AM)

You have a point. Why can't I type it out the way its supposed to look?

All pretty and confusing-like?



VaguelyCurious -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:55:03 AM)


quote:

ORIGINAL: MissAsylum

You have a point. Why can't I type it out the way its supposed to look?

All pretty and confusing-like?

If you've got a fairly recent version of microsoft word, it should have an equation builder. That's what you want.



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 12:01:33 PM)

I don't since my laptop is a dinosaur.



VaguelyCurious -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 12:13:15 PM)

Lance, I'm still getting a value that's half yours for that integer, and I don't know why.

The integer (if we're leaving everything inside that fourth root) is: (3^1/4 + 243^1/4 + 48^1/4)^4

243^1/4 = 3(3^(1/4))

48^1/4 = 2(3^(1/4))

so the part inside the 4th power is:

1(3^(1/4)) + 3(3^(1/4)) + 2(3^(1/4)) = 6(3^(1/4))

And when you ^4 that you get 3(6^4) = 3888

So I don't know why your figure is double mine.



MasterG2kTR -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 12:43:10 PM)

quote:

ORIGINAL: LanceHughes

6xy^2 could be just about anything!

inc. (6^2)(x^2)(y^2)


NO...it can't be that because that would be the same as 36x^2y^2

There are only three expressions in the final answer. (6) (x) & (y^2) the ^2 you see in the solution only applies to the y expression no the 6 & x

quote:

ORIGINAL: VaguelyCurious
And when you ^4 that you get 3(6^4) = 3888


Can't be that either because you have no idea what the value of x or y is. You can only end up with an exponential expression without a definitive solution until you know the actual values of x & y.


ETA: And don't forget you still have the 4th root of 3x as well.



shallowdeep -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 12:52:28 PM)

quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: shallowdeep
6 * x * y^2 * (3 * x)^.25

FAIL!

quote:

ORIGINAL: LanceHughes

X * (Y^2) * 6* (3X)^.25

Final answer.

Ah! I knew I shouldn't have trusted all those teachers when they claimed scalar multiplication was commutative!

Or, if I ask very nicely, do you suppose there might be a chance I could get my grade revised? :)

Of course, I happily admit your pedagogical technique was far better… I was just being impatient.

quote:

ORIGINAL: VaguelyCurious
Lance, I'm still getting a value that's half yours for that integer, and I don't know why.

Lance accidentally started assuming a 6 inside the fourth root, rather than a 3, in the post where he pointed out your mistake. The correct value is 3888. As someone may have mentioned…



LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 1:10:00 PM)

quote:

ORIGINAL: MasterG2kTR

quote:

ORIGINAL: LanceHughes

6xy^2 could be just about anything!

inc. (6^2)(x^2)(y^2)


NO...it can't be that because that would be the same as 36x^2y^2

There are only three expressions in the final answer. (6) (x) & (y^2) the ^2 you see in the solution only applies to the y expression not to the 6 & x

<snipped>

Well, that's my point. HOW is the reader to KNOW the span of the squaring symbol.  Maybe we could compromise on:  6X * Y^2



LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 1:16:36 PM)

quote:

ORIGINAL: shallowdeep

quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: shallowdeep
6 * x * y^2 * (3 * x)^.25

FAIL!

quote:

ORIGINAL: LanceHughes

X * (Y^2) * 6* (3X)^.25

Final answer.

Ah! I knew I shouldn't have trusted all those teachers when they claimed scalar multiplication was commutative!

Or, if I ask very nicely, do you suppose there might be a chance I could get my grade revised? :)

Of course, I happily admit your pedagogical technique was far better… I was just being impatient.

HUGE BLUSH! ! ! !  I guess I was seeing (or not seeing) parenthesisessissippi.

quote:

ORIGINAL: VaguelyCurious
Lance, I'm still getting a value that's half yours for that integer, and I don't know why.

Lance accidentally started assuming a 6 inside the fourth root, rather than a 3, in the post where he pointed out your mistake. The correct value is 3888. As someone may have mentioned…

HUGE(r) BLUSH! ! ! ! 



MasterG2kTR -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 1:21:13 PM)

quote:

ORIGINAL: LanceHughes

Well, that's my point. HOW is the reader to KNOW the span of the squaring symbol.  Maybe we could compromise on:  6X * Y^2



If the expression was listed like this (6xy)^2 then the square would apply to all three expressions and your assumption would be correct. Any power indicator only applies to the expression it is directly related to unless shown as above.



LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 1:31:16 PM)

Another! Another!  Give us another, Teacher!  Pleeeeeeease!

// G'damn, f*ckin' Lance //



MasterG2kTR -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 1:35:21 PM)

*looks for ruler*

Hold out those knuckles young man

(you know what's coming next I assume)

Yup....trip to the principle's office



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