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Question regarding quadratic equation roots watch

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    (Original post by SeanFM)
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    I know, I'm sorry lol. I keep doing that, I just don't understand how you factorise it.

    EDIT: My friend kind of helped me out here.

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

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

x-1[x^2-3x]=0

(x-1)(x)(x-3) = 0
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    (Original post by Chittesh14)
    I know, I'm sorry lol. I keep doing that, I just don't understand how you factorise it.
    (x-1)^3 - (x-1)(x+1) = 0.

    If we call (x-1) a and (x+1) b then you have a^3 - ab = 0.

    So how do you factorise that?
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    It turns into a(a^2 - b). If you can't see how, multiply that out.

    Then put it back in then you have (x-1)((x-1)^2 - (x+1)) ... and then all of the lines from your V2 follow.
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    (Original post by SeanFM)
    (x-1)^3 - (x-1)(x+1) = 0.

    If we call (x-1) a and (x+1) b then you have a^3 - ab = 0.

    So how do you factorise that?
    Spoiler:
    Show
    It turns into a(a^2 - b). If you can't see how, multiply that out.
    Then put it back in then you have (x-1)((x-1)^2 - (x+1)) ... and then all of the lines from your V2 follow.
    I've got it, thanks buddy!
    I don't know why, sometimes, my brain just can't do this... lol
    Sometimes, I just forget how to factorise equations which are raised to a power greater than 2.
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    (Original post by Chittesh14)
    I've got it, thanks buddy!
    I don't know why, sometimes, my brain just can't do this... lol
    Sometimes, I just forget how to factorise equations which are raised to a power greater than 2.
    It's okay, it may be tough at first but you've seen it done now so hopefully you should be okay :borat:
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    (Original post by SeanFM)
    It's okay, it may be tough at first but you've seen it done now so hopefully you should be okay :borat:
    No, it's actually easy lol.
    It's just that I haven't done it for a long time lol.
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    (Original post by Chittesh14)
    No, it's actually easy lol.
    It's just that I haven't done it for a long time lol.
    I was trying to make you feel better

    Don't be too hard on yourself
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    (Original post by SeanFM)
    I was trying to make you feel better

    Don't be too hard on yourself
    Oh right lol, thanks .

    Here's another question .

    x(x-2) = x(x-2)(x-3)

    I can't factorise here right?
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    (Original post by Chittesh14)
    Oh right lol, thanks .

    Here's another question .

    x(x-2) = x(x-2)(x-3)

    I can't factorise here right?
    Yes you can.

    x(x-2) - x(x-2)(x-3) = 0, so..

    Whenever you want to 'cancel out' something from both sides, like x(x-2) = x(x-2)(x-3)
    those are the things you need to factor out.
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    (Original post by SeanFM)
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    x(x-2)[1-(x-3)] = 0

x(x-2)[-x+4] = 0

x(x-2)(-x+4) = 0

x(x-2)(x-4) = 0

    Lol, why do I always get it wrong when I do it in my book. I just can't think of applying this same principle lol.
    Well, I've learnt something new !

    I hope this is right .
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    (Original post by Chittesh14)
    x(x-2)[1-(x-3)] = 0

x(x-2)[-x+4] = 0

x(x-2)(-x+4) = 0

x(x-2)(x-4) = 0

    Lol, why do I always get it wrong when I do it in my book. I just can't think of applying this same principle lol.
    Well, I've learnt something new !

    I hope this is right .
    Correct. Well done :borat:

    So just think, like a^2 + ab = a(a+b) and a+ ab = a(1 +b) etc.
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    (Original post by SeanFM)
    Correct. Well done :borat:

    So just think, like a^2 + ab = a(a+b) and a+ ab = a(1 +b) etc.
    Yeah, I'll try to remember it in that way, or a way that I can remember it .
    Thank you so much though :P
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    Hey, I keep making mistakes when drawing curves.

    This graph is x^3 - 3x^2 - 4x = 0

    Why can't I draw the graph in the way I've drawn it in the second picture?
    Is there something that will tell me that the curve goes downwards and not upwards after it crosses the y-axis? If so, what is it?
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