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Help with Sketching curves/Simultaneous equation

  1. Mar 22, 2013 #1
    Hi, I'm stuck on what should be an easy problem and I'm not sure where I'm going wrong. I already know the answer (I'm self studying), it's the method for the solution I don't understand. Here's the problem:

    a) on the same axes sketch the curves given by $$y= (x-1)^3$$ and
    $$y= (x-1)(1+x)$$

    b) find the coordinates of the points of intersection

    So, I did part a, and I thought i did part b, but I only got 1 of 3 possible solutions.

    Anyway, here was my attempt at a solution:
    $$ (x-1)^3 = (x-1)(1+x)$$
    $$(x-1)^2 = (1+x)$$
    Then multiply out the brackets: $$x^2 - 2x + 1 = 1+x$$
    Simplify: $$x^2 - 3x = 0$$
    Factorise: $$x(x-3) = 0$$

    From this I can get 2 values for x (0 and 3) and plug them into the original equations. I know these two are correct, but I'm missing one and I'm pretty sure my method is wrong.

    I had a look at the solution (the textbook I have comes with online worked solutions) but I've been staring at it and don't understand it. the solution they give is:

    $$(x-1)^3 = (x-1)(x+1)$$
    $$(x-1)^3 - (x-1)(x+1) = 0$$

    This is the part I don't understand at all:

    $$(x-1)[x^2-2x+1-(x+1)] = 0$$

    Then $$(x-1)(x^2-3x)=0$$
    $$(x-1)(x-3)x = 0$$
    $$x = 0, 1, 3$$

    Any help on this would be greatly appreciated, thank you.
     
  2. jcsd
  3. Mar 22, 2013 #2

    BruceW

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    Yeah, you've done good, but you just missed one of the points. You had this equation:
    [tex](x-1)^3 = (x-1)(1+x)[/tex]
    And then after this, you divided by (x-1) on both sides. This is the correct method to get two of the points. But before you do the division, you must identify the other point. Another way to think of it is - can you always divide by (x-1) ? What are you assuming here?
     
  4. Mar 22, 2013 #3
    EDIT - just realised all the working in my reply was wrong. Maybe I'm just having a bad day but I'm not able to get this at all... What am I assuming?!
     
  5. Mar 22, 2013 #4

    BruceW

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    your reply? did you delete your reply? I don't see it...

    Anyway, so to divide by (x-1), what value of x will give you problems?
     
  6. Mar 22, 2013 #5
    Yeah I deleted it, I didn't want you to waste your time replying to something I knew was incorrect.

    I think I understand now, x cannot be 1 can it? If it is then you're dividing by 0 which means the answer will be undefined, but if x can't be 1 then how does it end up being one of the coordinates? Looking at the equation, I can see that subbing in 1 for x does work, I'm just not sure how I would come to that conclusion methodically. Does the method that I showed in my original post, the one given in the solutions book, make sense to you?
     
  7. Mar 24, 2013 #6

    BruceW

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    yeah, exactly. You can't divide by (x-1) when x has the value one. And yes, subbing in for x=1 does work, so x=1 must be a solution. You can think of it like this: OK, you've found that x=1 is one solution. So then to find other solutions, it is OK to divide by (x-1) because for other solutions, x must be different from one.

    And the solutions book method is correct. It is pretty much the same as your method, they just did it slightly differently. The way they did it, they rearranged things so that they ended up with the three factors. So then at the end they have the nice equation:
    [tex](x-1)(x-3)x = 0[/tex]
    Where the three solutions are all in one line. But your method was also correct. It is just that when you got to the step where you divided by (x-1), you should have written "and here, you can see x=1 is one solution" or something like that, and then continued to find the other two solutions.
     
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