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Envii
Please help.
what I think you should do is first multiply the (x-1) by (x^2-4)
then square both sides, and then collect the like terms
How do I multiply (x-1) by (x^2-4)?
Yay. :D Then how do I combine like terms?
you have to square both sides first, then collect the like terms, then isolate x
there might be an easier way, but i'm not sure
to get rid of the sqrt sign, you'll need to square both sides: \[(x^3-x^2-4x+4)^2=x^4-2x^3\]
Wait, wouldn't it be (x^4-2x^3)^2 on that side?
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\[a^1=a\]
Do I have to do the (x^3-x^2-4x+4)^2 first?
yes, now it's a very, very, very long process. Are you allowed to use a calculator for this part?
so type \[(x^3 −x^2 −4x+4)^2\] into the calculator to get an answer
How do I do that if I'm using the google calculator?
I've never used a google calc. let me see
The calculator I normally use is broken, sorry. :(
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
Alright I got it. What next?
subtract x^4 from both sides and add 2x^3 to get one side equal to 0
Alright, now I can do x^2-x^2 to get -, correct?
but, yes, use that idea to get all the terms on one side
x^6-x^4=x^2 and -2x^5+2x^3=x^2, right?
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
so what you should have gotten is: \[x^6-2x^5-7x^4+16x^3+8x^2-32x+16=x^4+2x^3\]
Mind if I make a suggestion that will keep things simpler?
@whpalmer4 of course @Chelsea04 Wouldn't it be =x^4-2x^3?
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 \[(x-1) = \frac{\sqrt{x^4-2x^3}}{(x+2)(x-2)} = \frac{\sqrt{x^3(x-2)}}{(x+2)(x-2)} = \frac{x\sqrt{x(x-2)}}{(x+2)(x-2)}\]
yea, but how does that help? I noticed it too, but didn't know what to do with it
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\[(x-1)^2(x+2)^2(x-2) = x^3\]
we marvel at how beautiful it is? :-)
wouldn't you have to eventually expand it to solve for x? @whpalmer4 that's funny, lol :P
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...
If we expand it out, we get \[x^5-8x^3+2x^2+12x-8=0\]
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: \[\{\{x\to -2.53629\},\{x\to -1.59124\},\{x\to 0.8782\, -0.254473 i\},\]\[\{x\to 0.8782\, +0.254473 i\},\{x\to 2.37113\}\}\]
I am just so confused right now.
what exactly does the question ask?
I guess you could've just typed the whole thing into the calculator
Now here's a thought, what if the problem is really \[x-1 = \sqrt{\frac{x^4-2x^3}{x^2-4} }\]
that'd make more sense
Then we have \[(x-1)^2 = \frac{x^3(x-2)}{(x-2)(x+2)}\]
is that the question @Envii ?
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\[x^3+2x^2-2x^2-4x+x+2 = x^3\]
@Envil, why don't you try solving that?
This version is MUCH easier :-)
YUP!!! MUCH MUCH EASIER!!!
I did like the look of the other one in factored form, though :-)
@Envil have you simplified and solved that equation yet?
Working on it. Sorry, I suck at math. xD
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?
Right. And what is -4x + x?
Come on, you're really close to the answer!
\[-3x+2 = 0\]Add 3x to both sides, and then divide both sides by 3 to get x = ????
I'd keep it in exact form: x = 2/3
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.
Well, did I lose you in the painful fraction arithmetic, or in the decision that the solution wasn't valid?
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!