Please help.

- anonymous

Please help.

- jamiebookeater

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- anonymous

|dw:1362030447502:dw|

- anonymous

what I think you should do is first multiply the (x-1) by (x^2-4)

- anonymous

then square both sides, and then collect the like terms

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## More answers

- anonymous

How do I multiply (x-1) by (x^2-4)?

- anonymous

expand the brackets

- anonymous

Is it x^3-x^2-4x+4?

- anonymous

yea

- anonymous

Yay. :D Then how do I combine like terms?

- anonymous

you have to square both sides first, then collect the like terms, then isolate x

- anonymous

there might be an easier way, but i'm not sure

- anonymous

Square both sides? How?

- anonymous

to get rid of the sqrt sign, you'll need to square both sides:
\[(x^3-x^2-4x+4)^2=x^4-2x^3\]

- anonymous

Wait, wouldn't it be (x^4-2x^3)^2 on that side?

- anonymous

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\[a^1=a\]

- anonymous

Oh I get it.

- anonymous

Do I have to do the (x^3-x^2-4x+4)^2 first?

- anonymous

yes, now it's a very, very, very long process. Are you allowed to use a calculator for this part?

- anonymous

Yes I am.

- anonymous

so type \[(x^3 âˆ’x^2 âˆ’4x+4)^2\] into the calculator to get an answer

- anonymous

How do I do that if I'm using the google calculator?

- anonymous

I've never used a google calc. let me see

- anonymous

The calculator I normally use is broken, sorry. :(

- anonymous

try use this site,
http://www.wolframalpha.com/input/?i=%28x%5E3-x%5E2-4x%2B4%29%5E2
I've already typed in the equation, look for the words expanded form

- anonymous

Alright I got it. What next?

- anonymous

subtract x^4 from both sides and add 2x^3 to get one side equal to 0

- anonymous

Alright, now I can do x^2-x^2 to get -, correct?

- anonymous

to get 0*

- anonymous

how'd you get x^2?

- anonymous

but, yes, use that idea to get all the terms on one side

- anonymous

x^6-x^4=x^2 and -2x^5+2x^3=x^2, right?

- anonymous

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

- anonymous

so what you should have gotten is: \[x^6-2x^5-7x^4+16x^3+8x^2-32x+16=x^4+2x^3\]

- whpalmer4

Mind if I make a suggestion that will keep things simpler?

- anonymous

@whpalmer4 of course @Chelsea04 Wouldn't it be =x^4-2x^3?

- anonymous

oh, yes sorry.

- whpalmer4

Way back at the beginning, notice that \(x^2-4\) is the difference of two squares, and factors to \((x+2)(x-2)\)
We can write the problem as
\[(x-1) = \frac{\sqrt{x^4-2x^3}}{(x+2)(x-2)} = \frac{\sqrt{x^3(x-2)}}{(x+2)(x-2)} = \frac{x\sqrt{x(x-2)}}{(x+2)(x-2)}\]

- anonymous

yea, but how does that help? I noticed it too, but didn't know what to do with it

- whpalmer4

Now if we keep everything like that, and multiply both sides by the denominator, we get
\[(x-1)(x+2)(x-2) = x\sqrt{x(x-2)}\]and after we square both sides, we just tack on exponents and get rid of the radical sign:
\[(x-1)^2(x+2)^2(x-2)^2 = x^2x(x-2)\]and now we can cancel out a common term from both sides\[(x-1)^2(x+2)^2(x-2) = x^3\]

- anonymous

then what?

- whpalmer4

we marvel at how beautiful it is? :-)

- anonymous

wouldn't you have to eventually expand it to solve for x?
@whpalmer4 that's funny, lol :P

- whpalmer4

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...

- whpalmer4

If we expand it out, we get
\[x^5-8x^3+2x^2+12x-8=0\]

- whpalmer4

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:
\[\{\{x\to -2.53629\},\{x\to -1.59124\},\{x\to 0.8782\, -0.254473 i\},\]\[\{x\to 0.8782\, +0.254473 i\},\{x\to 2.37113\}\}\]

- anonymous

I am just so confused right now.

- anonymous

what exactly does the question ask?

- anonymous

It just says solve.

- anonymous

I guess you could've just typed the whole thing into the calculator

- whpalmer4

Now here's a thought, what if the problem is really
\[x-1 = \sqrt{\frac{x^4-2x^3}{x^2-4} }\]

- anonymous

that'd make more sense

- whpalmer4

Then we have
\[(x-1)^2 = \frac{x^3(x-2)}{(x-2)(x+2)}\]

- anonymous

is that the question @Envii ?

- whpalmer4

Cancel out (x-2) and multiply both sides by (x+2) gives us
\[(x-1)^2(x+2) = x^3\]Expand the left side
\[(x^2-2x+1)(x+2) = x^3\]Expand again\[x^3+2x^2-2x^2-4x+x+2 = x^3\]

- whpalmer4

@Envil, why don't you try solving that?

- whpalmer4

This version is MUCH easier :-)

- anonymous

YUP!!! MUCH MUCH EASIER!!!

- whpalmer4

I did like the look of the other one in factored form, though :-)

- whpalmer4

@Envil have you simplified and solved that equation yet?

- anonymous

Working on it. Sorry, I suck at math. xD

- whpalmer4

Here's the equation again:
\[x^3+2x^2-2x^2-4x+x+2 = x^3\]What do you have left after you collect like terms?

- anonymous

-4x+x+2=0

- whpalmer4

Right. And what is -4x + x?

- whpalmer4

Come on, you're really close to the answer!

- anonymous

-3x+2.

- whpalmer4

\[-3x+2 = 0\]Add 3x to both sides, and then divide both sides by 3 to get x = ????

- anonymous

x=0.66?

- whpalmer4

I'd keep it in exact form: x = 2/3

- whpalmer4

But now we need to check it, because we squared both sides and that sometimes introduces fictitious solutions.
x-1 = 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,
\[-\frac{1}3 = \sqrt{\frac{(\frac{2}{3})^4-2(\frac{2}{3})^3}{(\frac{2}{3})^2 -4}} = \sqrt{\frac{(\frac{16}{81})-2(\frac{8}{27})}{(\frac{4}{9}) -4}} = \sqrt{\frac{(\frac{16}{81})-(\frac{48}{81})}{(\frac{4}{9}) -(\frac{36}{9})}} = \sqrt{\frac{\frac{-32}{81}}{\frac{-32}{9}}}=\sqrt{\frac{1}9}\]
So it isn't a solution after all.

- anonymous

You lost me. D:

- whpalmer4

Well, did I lose you in the painful fraction arithmetic, or in the decision that the solution wasn't valid?

- whpalmer4

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:
\[1-x = \sqrt{\frac{x^4-2x^3}{x^2-4}}\]We proceed the same way:
\[(1-x)^2(x+2)(x-2) = x^3(x-2)\]Cancel common factors\[(1-x)^2(x+2)=x^3\]Expand\[(1-2x+x^2)(x+2)=x^3\]Expand again\[x+2-2x^2-4x+x^3+2x^2=x^3\]Collect like terms\[-3x+2=0\]\[x=\frac{2}3\]
If we plug that back into its original equation:
\[1-\frac{2}{3} = \sqrt{\frac{(\frac{2}{3})^4-2(\frac{2}{3})^3}{(\frac{2}{3})^2-4}} \]\[\frac{1}3 = \sqrt{\frac{1}{9}}\]and that is true...I wonder if that was the correct problem!

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