1. Envii Group Title

|dw:1362030447502:dw|

2. Chelsea04 Group Title

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

3. Chelsea04 Group Title

then square both sides, and then collect the like terms

4. Envii Group Title

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

5. Chelsea04 Group Title

expand the brackets

6. Envii Group Title

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

7. Chelsea04 Group Title

yea

8. Envii Group Title

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

9. Chelsea04 Group Title

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

10. Chelsea04 Group Title

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

11. Envii Group Title

Square both sides? How?

12. Chelsea04 Group Title

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 Group Title

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

14. Chelsea04 Group Title

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 Group Title

Oh I get it.

16. Envii Group Title

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

17. Chelsea04 Group Title

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

18. Envii Group Title

Yes I am.

19. Chelsea04 Group Title

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

20. Envii Group Title

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

21. Chelsea04 Group Title

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

22. Envii Group Title

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

23. Chelsea04 Group Title

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 Group Title

Alright I got it. What next?

25. Chelsea04 Group Title

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

26. Envii Group Title

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

27. Envii Group Title

to get 0*

28. Chelsea04 Group Title

how'd you get x^2?

29. Chelsea04 Group Title

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

30. Envii Group Title

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

31. Chelsea04 Group Title

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 Group Title

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 Group Title

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

34. Envii Group Title

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

35. Chelsea04 Group Title

oh, yes sorry.

36. whpalmer4 Group Title

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 Group Title

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

38. whpalmer4 Group Title

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 Group Title

then what?

40. whpalmer4 Group Title

we marvel at how beautiful it is? :-)

41. Chelsea04 Group Title

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

42. whpalmer4 Group Title

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 Group Title

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

44. whpalmer4 Group Title

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 Group Title

I am just so confused right now.

46. Chelsea04 Group Title

what exactly does the question ask?

47. Envii Group Title

It just says solve.

48. Chelsea04 Group Title

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

49. whpalmer4 Group Title

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

50. Chelsea04 Group Title

that'd make more sense

51. whpalmer4 Group Title

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

52. Chelsea04 Group Title

is that the question @Envii ?

53. whpalmer4 Group Title

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 Group Title

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

55. whpalmer4 Group Title

This version is MUCH easier :-)

56. Chelsea04 Group Title

YUP!!! MUCH MUCH EASIER!!!

57. whpalmer4 Group Title

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

58. whpalmer4 Group Title

@Envil have you simplified and solved that equation yet?

59. Envii Group Title

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

60. whpalmer4 Group Title

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 Group Title

-4x+x+2=0

62. whpalmer4 Group Title

Right. And what is -4x + x?

63. whpalmer4 Group Title

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

64. Envii Group Title

-3x+2.

65. whpalmer4 Group Title

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

66. Envii Group Title

x=0.66?

67. whpalmer4 Group Title

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

68. whpalmer4 Group Title

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 Group Title

You lost me. D:

70. whpalmer4 Group Title

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

71. whpalmer4 Group Title

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!