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

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MissAsylum -> Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 7:18:45 AM)

Word-for-word problem

Combine the expressions. (Assume the variables are nonnegative numbers)

x^4√3xy^8 + y^4√243x^5y^4 + y^2 4^√48x^5

i'm stumpped 3 ways til sunday. Can anybody help?



MadAxeman -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 7:37:58 AM)

Isn't that the Binomial Expansion Theorem?

If you google Pascal's Triangle there should be a comprehensive deconstruction



SoulAlloy -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 9:51:33 AM)

Where do the "power" and "square route" symbols end?

For example the middle section y^4√243x^5y^4 - is it y to the power of 4√243x times x to the power of 5y to the power of 4?

Or is it y to the power of 4 times the square route of 243x to the power of 5 times y to the power of 5?

There's other combinations from reading it too, a bit hard to combine without knowing the beginning/ends



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 10:02:04 AM)


quote:

ORIGINAL: SoulAlloy

Where do the "power" and "square route" symbols end?

For example the middle section y^4√243x^5y^4 - is it y to the power of 4√243x times x to the power of 5y to the power of 4?

Or is it y to the power of 4 times the square route of 243x to the power of 5 times y to the power of 5?

There's other combinations from reading it too, a bit hard to combine without knowing the beginning/ends


Its the second possibility that you mentioned.




LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 10:19:49 AM)

quote:

ORIGINAL: MissAsylum

Word-for-word problem

Combine the expressions. (Assume the variables are nonnegative numbers)

x^4√3xy^8 + y^4√243x^5y^4 + y^2 4^√48x^5

i'm stumpped 3 ways til sunday. Can anybody help?

PARENTHESIS, please!  So far, (I'm guessing) we have:

x^4√3xy^8 + (y^4)(√(243x)^5)(y^4) + y^2 4^√48x^5

So, the FIRST thing that jumps out of the middle term is (y^4)(y^4) = y^8, so the first term and second have at least one common factor.  BUT!

What is the subject at hand?  Factoring trinomials?  The subject material WILL give a HUGE clue.

We have:

x^4√3xy^8 + (y^8)(√(243x)^5) + y^2 4^√48x^5

Same idea with
(x^4)x in first term = x^5 in both 2nd and third terms.  BUT!  Really, really need to know where pieces begin end for middle term.  I'm hoping the radical sign does NOT cover x^5 in second term.  If it doesn't we can write (x^5)x(bunch of stuff) and then see if that y term might break-down.  I suspect (x^5 ) x {y^2  x [√48+y^6 x (√(243) ] }  or.... something like that.  One term got lost while I was trying to get []'s and {}'s lined up.... and still need parens.  Maybe re-write with numerical roots at back of each term... that is, if they are just numerical roots.



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

Well the first two terms simplify down to (x^5)(y^8)(10√3) pretty easily, but I have no idea what the third term is supposed to mean. Write it out using actual brackets? Because I suspect it's also supposed to be in terms of (X^5)(y^8) and you've just typed it out wrong.



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 10:40:39 AM)

error.




mnottertail -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 10:41:54 AM)

you need to post reply, load the picture and embed in post.



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 10:42:23 AM)

Photo of the problem

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



LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 10:53:37 AM)

YIKES!  Those are fourth roots!  Square roots of square roots!
First term: X * {3 * X * (Y^8)}^ (1/4)
First term: X * {[3 * X]^ (1/4)} * { (Y^8)^ (1/4)} because any root of a multiplicative expression is the same root of each part multiplied
First term: X * {[3 * X]^ (1/4)} * {  Y^2 } because the quarter root of an eight power is a square root.
First term: X(Y^2) * {[3X]^ (1/4)}

Second term: Y * {243 * X^5 * Y^4} ^ (1/4) That 4th root of three makes me wonder about 243.  243 / 3 = 81 * 3, so we write:
Second term: Y * {3^5 * X^5 * Y^4} ^ (1/4)
Second term: Y * {[3^4 * X^4 * Y^4] * [3 * X]} ^ (1/4)
Second term: Y * {[3^4 * X^4 * Y^4] ^ (1/4)}  * {[3 * X] ^ (1/4)}
Second term: Y * {3 * X * Y} * {[3 * X] ^ (1/4)}
Second term: 3 X( Y^2) * {[3X] ^ (1/4)}

Third term:  Left for the student



VaguelyCurious -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:06:06 AM)

quote:

ORIGINAL: MissAsylum

Photo of the problem

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

Ah. Then ignore what I just said about the first two terms, ok? Brb.

ETA: also, that bears absolutely no relation to what you actually typed out. You can see that, right?



shallowdeep -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:08:58 AM)

6 * x * y^2 * (3 * x)^.25

or, equivalently,

(3888 * x^5 * y^8)^.25



VaguelyCurious -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:10:59 AM)

I love you, man. Really I do. [:D]



LanceHughes -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:13:40 AM)

quote:

ORIGINAL: shallowdeep

6 * x * y^2 * (3 * x)^.25

or, equivalently,

(3888 * x^5 * y^8)^.25

FAIL!

ETA:  There are no factors of 6 across the three original terms. Not even factors of 2. 



VaguelyCurious -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:16:50 AM)

Although I disagree with you. I got 294^0.25 as the integer. Where'd you get such a big number?



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:18:11 AM)


quote:

ORIGINAL: VaguelyCurious

quote:

ORIGINAL: MissAsylum

Photo of the problem

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

Ah. Then ignore what I just said about the first two terms, ok? Brb.

ETA: also, that bears absolutely no relation to what you actually typed out. You can see that, right?


I tried. Isn't that the most important thing? Lol



VaguelyCurious -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:19:37 AM)

Not in maths, honey. Do you need the answer explaining, or do you just need to know what it is?



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:20:05 AM)


quote:

ORIGINAL: LanceHughes

YIKES!  Those are fourth roots!  Square roots of square roots!
First term: X * {3 * X * (Y^8)}^ (1/4)
First term: X * {[3 * X]^ (1/4)} * { (Y^8)^ (1/4)} because any root of a multiplicative expression is the same root of each part multiplied
First term: X * {[3 * X]^ (1/4)} * {  Y^2 } because the quarter root of an eight power is a square root.
First term: X(Y^2) * {[3X]^ (1/4)}

Second term: Y * {243 * X^5 * Y^4} ^ (1/4) That 4th root of three makes me wonder about 243.  243 / 3 = 81 * 3, so we write:
Second term: Y * {3^5 * X^5 * Y^4} ^ (1/4)
Second term: Y * {[3^4 * X^4 * Y^4] * [3 * X]} ^ (1/4)
Second term: Y * {[3^4 * X^4 * Y^4] ^ (1/4)}  * {[3 * X] ^ (1/4)}
Second term: Y * {3 * X * Y} * {[3 * X] ^ (1/4)}
Second term: 3 X( Y^2) * {[3X] ^ (1/4)}

Third term:  Left for the student


Thank you! Now I can actually work out the problem!



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:22:05 AM)


quote:

ORIGINAL: LanceHughes

quote:

ORIGINAL: shallowdeep

6 * x * y^2 * (3 * x)^.25

or, equivalently,

(3888 * x^5 * y^8)^.25

FAIL!

ETA:  There are no factors of 6 across the three original terms. Not even factors of 2. 


I have the mathematical ability of a 3 year old, and even I knew that wasn't right.



MissAsylum -> RE: Any Math Wizards Care to Take A Crack At This Problem? (10/12/2011 11:23:59 AM)

So.... 6xy^2 4^√3x.


I think.



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