anonymous
  • anonymous
Please help.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1362030447502:dw|
anonymous
  • anonymous
what I think you should do is first multiply the (x-1) by (x^2-4)
anonymous
  • anonymous
then square both sides, and then collect the like terms

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More answers

anonymous
  • anonymous
How do I multiply (x-1) by (x^2-4)?
anonymous
  • anonymous
expand the brackets
anonymous
  • anonymous
Is it x^3-x^2-4x+4?
anonymous
  • anonymous
yea
anonymous
  • anonymous
Yay. :D Then how do I combine like terms?
anonymous
  • anonymous
you have to square both sides first, then collect the like terms, then isolate x
anonymous
  • anonymous
there might be an easier way, but i'm not sure
anonymous
  • anonymous
Square both sides? How?
anonymous
  • 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
  • anonymous
Wait, wouldn't it be (x^4-2x^3)^2 on that side?
anonymous
  • 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
  • anonymous
Oh I get it.
anonymous
  • anonymous
Do I have to do the (x^3-x^2-4x+4)^2 first?
anonymous
  • anonymous
yes, now it's a very, very, very long process. Are you allowed to use a calculator for this part?
anonymous
  • anonymous
Yes I am.
anonymous
  • anonymous
so type \[(x^3 −x^2 −4x+4)^2\] into the calculator to get an answer
anonymous
  • anonymous
How do I do that if I'm using the google calculator?
anonymous
  • anonymous
I've never used a google calc. let me see
anonymous
  • anonymous
The calculator I normally use is broken, sorry. :(
anonymous
  • 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
  • anonymous
Alright I got it. What next?
anonymous
  • anonymous
subtract x^4 from both sides and add 2x^3 to get one side equal to 0
anonymous
  • anonymous
Alright, now I can do x^2-x^2 to get -, correct?
anonymous
  • anonymous
to get 0*
anonymous
  • anonymous
how'd you get x^2?
anonymous
  • anonymous
but, yes, use that idea to get all the terms on one side
anonymous
  • anonymous
x^6-x^4=x^2 and -2x^5+2x^3=x^2, right?
anonymous
  • 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
  • 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
  • whpalmer4
Mind if I make a suggestion that will keep things simpler?
anonymous
  • anonymous
@whpalmer4 of course @Chelsea04 Wouldn't it be =x^4-2x^3?
anonymous
  • anonymous
oh, yes sorry.
whpalmer4
  • 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
  • anonymous
yea, but how does that help? I noticed it too, but didn't know what to do with it
whpalmer4
  • 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
  • anonymous
then what?
whpalmer4
  • whpalmer4
we marvel at how beautiful it is? :-)
anonymous
  • anonymous
wouldn't you have to eventually expand it to solve for x? @whpalmer4 that's funny, lol :P
whpalmer4
  • 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
  • whpalmer4
If we expand it out, we get \[x^5-8x^3+2x^2+12x-8=0\]
whpalmer4
  • 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
  • anonymous
I am just so confused right now.
anonymous
  • anonymous
what exactly does the question ask?
anonymous
  • anonymous
It just says solve.
anonymous
  • anonymous
I guess you could've just typed the whole thing into the calculator
whpalmer4
  • whpalmer4
Now here's a thought, what if the problem is really \[x-1 = \sqrt{\frac{x^4-2x^3}{x^2-4} }\]
anonymous
  • anonymous
that'd make more sense
whpalmer4
  • whpalmer4
Then we have \[(x-1)^2 = \frac{x^3(x-2)}{(x-2)(x+2)}\]
anonymous
  • anonymous
is that the question @Envii ?
whpalmer4
  • 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
  • whpalmer4
@Envil, why don't you try solving that?
whpalmer4
  • whpalmer4
This version is MUCH easier :-)
anonymous
  • anonymous
YUP!!! MUCH MUCH EASIER!!!
whpalmer4
  • whpalmer4
I did like the look of the other one in factored form, though :-)
whpalmer4
  • whpalmer4
@Envil have you simplified and solved that equation yet?
anonymous
  • anonymous
Working on it. Sorry, I suck at math. xD
whpalmer4
  • 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
  • anonymous
-4x+x+2=0
whpalmer4
  • whpalmer4
Right. And what is -4x + x?
whpalmer4
  • whpalmer4
Come on, you're really close to the answer!
anonymous
  • anonymous
-3x+2.
whpalmer4
  • whpalmer4
\[-3x+2 = 0\]Add 3x to both sides, and then divide both sides by 3 to get x = ????
anonymous
  • anonymous
x=0.66?
whpalmer4
  • whpalmer4
I'd keep it in exact form: x = 2/3
whpalmer4
  • 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
  • anonymous
You lost me. D:
whpalmer4
  • whpalmer4
Well, did I lose you in the painful fraction arithmetic, or in the decision that the solution wasn't valid?
whpalmer4
  • 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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