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Envii Group TitleBest ResponseYou've already chosen the best response.0
dw:1362030447502:dw
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
what I think you should do is first multiply the (x1) by (x^24)
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
then square both sides, and then collect the like terms
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
How do I multiply (x1) by (x^24)?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
expand the brackets
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
Is it x^3x^24x+4?
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
Yay. :D Then how do I combine like terms?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
you have to square both sides first, then collect the like terms, then isolate x
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
there might be an easier way, but i'm not sure
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
Square both sides? How?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
to get rid of the sqrt sign, you'll need to square both sides: \[(x^3x^24x+4)^2=x^42x^3\]
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
Wait, wouldn't it be (x^42x^3)^2 on that side?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
no, because a sqrt and a squared cancels out. \[(\sqrt{a})^2=a\]because\[\sqrt{a}=a^\frac{ 1 }{ 2 }\]and\[(\sqrt{a})^2=a^{\frac{ 1 }{ 2 }*2}\]which is\[a^1=a\]
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
Oh I get it.
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
Do I have to do the (x^3x^24x+4)^2 first?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
yes, now it's a very, very, very long process. Are you allowed to use a calculator for this part?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
so type \[(x^3 −x^2 −4x+4)^2\] into the calculator to get an answer
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
How do I do that if I'm using the google calculator?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
I've never used a google calc. let me see
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
The calculator I normally use is broken, sorry. :(
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
try use this site, http://www.wolframalpha.com/input/?i=%28x%5E3x%5E24x%2B4%29%5E2 I've already typed in the equation, look for the words expanded form
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
Alright I got it. What next?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
subtract x^4 from both sides and add 2x^3 to get one side equal to 0
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
Alright, now I can do x^2x^2 to get , correct?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
how'd you get x^2?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
but, yes, use that idea to get all the terms on one side
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
x^6x^4=x^2 and 2x^5+2x^3=x^2, right?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
no, you can't do that. they're not like terms. like terms are: x and 4x x^2 and 7x^2 not x^6 and x^4
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
so what you should have gotten is: \[x^62x^57x^4+16x^3+8x^232x+16=x^4+2x^3\]
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Mind if I make a suggestion that will keep things simpler?
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
@whpalmer4 of course @Chelsea04 Wouldn't it be =x^42x^3?
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
oh, yes sorry.
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Way back at the beginning, notice that \(x^24\) is the difference of two squares, and factors to \((x+2)(x2)\) We can write the problem as \[(x1) = \frac{\sqrt{x^42x^3}}{(x+2)(x2)} = \frac{\sqrt{x^3(x2)}}{(x+2)(x2)} = \frac{x\sqrt{x(x2)}}{(x+2)(x2)}\]
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
yea, but how does that help? I noticed it too, but didn't know what to do with it
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Now if we keep everything like that, and multiply both sides by the denominator, we get \[(x1)(x+2)(x2) = x\sqrt{x(x2)}\]and after we square both sides, we just tack on exponents and get rid of the radical sign: \[(x1)^2(x+2)^2(x2)^2 = x^2x(x2)\]and now we can cancel out a common term from both sides\[(x1)^2(x+2)^2(x2) = x^3\]
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
then what?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
we marvel at how beautiful it is? :)
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
wouldn't you have to eventually expand it to solve for x? @whpalmer4 that's funny, lol :P
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Are we 100% certain this was the problem to be solved, and we're really supposed to find the roots? We've got a 5th order polynomial here...
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
If we expand it out, we get \[x^58x^3+2x^2+12x8=0\]
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Can't factor it, and there's no exact general solution for 5th order or higher polynomials, so Mathematica just finds numeric values for this one: \[\{\{x\to 2.53629\},\{x\to 1.59124\},\{x\to 0.8782\, 0.254473 i\},\]\[\{x\to 0.8782\, +0.254473 i\},\{x\to 2.37113\}\}\]
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
I am just so confused right now.
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
what exactly does the question ask?
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
It just says solve.
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
I guess you could've just typed the whole thing into the calculator
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Now here's a thought, what if the problem is really \[x1 = \sqrt{\frac{x^42x^3}{x^24} }\]
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
that'd make more sense
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Then we have \[(x1)^2 = \frac{x^3(x2)}{(x2)(x+2)}\]
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
is that the question @Envii ?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Cancel out (x2) and multiply both sides by (x+2) gives us \[(x1)^2(x+2) = x^3\]Expand the left side \[(x^22x+1)(x+2) = x^3\]Expand again\[x^3+2x^22x^24x+x+2 = x^3\]
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
@Envil, why don't you try solving that?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
This version is MUCH easier :)
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
YUP!!! MUCH MUCH EASIER!!!
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
I did like the look of the other one in factored form, though :)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
@Envil have you simplified and solved that equation yet?
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
Working on it. Sorry, I suck at math. xD
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Here's the equation again: \[x^3+2x^22x^24x+x+2 = x^3\]What do you have left after you collect like terms?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Right. And what is 4x + x?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Come on, you're really close to the answer!
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
\[3x+2 = 0\]Add 3x to both sides, and then divide both sides by 3 to get x = ????
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
I'd keep it in exact form: x = 2/3
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
But now we need to check it, because we squared both sides and that sometimes introduces fictitious solutions. x1 = 2/3 1 = 1/3 Uh oh, that's trouble, because we've got a square root on the other side of the equals sign, which is going to give us a positive number. If we expand it out, \[\frac{1}3 = \sqrt{\frac{(\frac{2}{3})^42(\frac{2}{3})^3}{(\frac{2}{3})^2 4}} = \sqrt{\frac{(\frac{16}{81})2(\frac{8}{27})}{(\frac{4}{9}) 4}} = \sqrt{\frac{(\frac{16}{81})(\frac{48}{81})}{(\frac{4}{9}) (\frac{36}{9})}} = \sqrt{\frac{\frac{32}{81}}{\frac{32}{9}}}=\sqrt{\frac{1}9}\] So it isn't a solution after all.
 one year ago

Envii Group TitleBest ResponseYou've already chosen the best response.0
You lost me. D:
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
Well, did I lose you in the painful fraction arithmetic, or in the decision that the solution wasn't valid?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.2
After we plugged our "solution" back into the original equation and ground through the arithmetic, we ended up with this: \[\frac{1}{3} = \sqrt{\frac{1}9}\]Unfortunately, that isn't true. Therefore, our solution isn't valid. There's a similar problem that does have working solutions: \[1x = \sqrt{\frac{x^42x^3}{x^24}}\]We proceed the same way: \[(1x)^2(x+2)(x2) = x^3(x2)\]Cancel common factors\[(1x)^2(x+2)=x^3\]Expand\[(12x+x^2)(x+2)=x^3\]Expand again\[x+22x^24x+x^3+2x^2=x^3\]Collect like terms\[3x+2=0\]\[x=\frac{2}3\] If we plug that back into its original equation: \[1\frac{2}{3} = \sqrt{\frac{(\frac{2}{3})^42(\frac{2}{3})^3}{(\frac{2}{3})^24}} \]\[\frac{1}3 = \sqrt{\frac{1}{9}}\]and that is true...I wonder if that was the correct problem!
 one year ago
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