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1. Envii

|dw:1362030447502:dw|

2. Chelsea04

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

3. Chelsea04

then square both sides, and then collect the like terms

4. Envii

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

5. Chelsea04

expand the brackets

6. Envii

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

7. Chelsea04

yea

8. Envii

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

9. Chelsea04

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

10. Chelsea04

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

11. Envii

Square both sides? How?

12. Chelsea04

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

13. Envii

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

14. Chelsea04

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$a^1=a$

15. Envii

Oh I get it.

16. Envii

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

17. Chelsea04

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

18. Envii

Yes I am.

19. Chelsea04

so type $(x^3 −x^2 −4x+4)^2$ into the calculator to get an answer

20. Envii

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

21. Chelsea04

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

22. Envii

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

23. Chelsea04

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

24. Envii

Alright I got it. What next?

25. Chelsea04

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

26. Envii

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

27. Envii

to get 0*

28. Chelsea04

how'd you get x^2?

29. Chelsea04

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

30. Envii

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

31. Chelsea04

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

32. Chelsea04

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

33. whpalmer4

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

34. Envii

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

35. Chelsea04

oh, yes sorry.

36. 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 $(x-1) = \frac{\sqrt{x^4-2x^3}}{(x+2)(x-2)} = \frac{\sqrt{x^3(x-2)}}{(x+2)(x-2)} = \frac{x\sqrt{x(x-2)}}{(x+2)(x-2)}$

37. Chelsea04

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

38. 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$(x-1)^2(x+2)^2(x-2) = x^3$

39. Chelsea04

then what?

40. whpalmer4

we marvel at how beautiful it is? :-)

41. Chelsea04

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

42. 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...

43. whpalmer4

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

44. 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\}\}$

45. Envii

I am just so confused right now.

46. Chelsea04

what exactly does the question ask?

47. Envii

It just says solve.

48. Chelsea04

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

49. whpalmer4

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

50. Chelsea04

that'd make more sense

51. whpalmer4

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

52. Chelsea04

is that the question @Envii ?

53. 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$x^3+2x^2-2x^2-4x+x+2 = x^3$

54. whpalmer4

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

55. whpalmer4

This version is MUCH easier :-)

56. Chelsea04

YUP!!! MUCH MUCH EASIER!!!

57. whpalmer4

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

58. whpalmer4

@Envil have you simplified and solved that equation yet?

59. Envii

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

60. 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?

61. Envii

-4x+x+2=0

62. whpalmer4

Right. And what is -4x + x?

63. whpalmer4

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

64. Envii

-3x+2.

65. whpalmer4

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

66. Envii

x=0.66?

67. whpalmer4

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

68. 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.

69. Envii

You lost me. D:

70. whpalmer4

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

71. 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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