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A Tip for easily factorizing quadratics with complex roots but real coefficients watch

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    I thought that many people who are good at factorizing and don't like to use the quadratic formula would probably find this useful. This method is really easy, and once you grasp it you'll find those quadratics much easier to factorize than the ones with real roots but where the x^2 coefficient is not equal to 1. I'll put it in a step by step form, kind of, because that numbering thing isn't working for some reason.
    Make sure the coefficient of x^2 is positive!
    Also I will expect the trinomial to be put in the form ax^2+bx+c, where a>0, and I will be referring to the coefficients by their corresponding letters.

    Recognize the quadratic:
    By this I mean make sure the quadratic does have complex roots. An easy way to see that it does NOT is when c is negative. Also when c is positive and large compared to b it probably has complex roots.

    When a=1:
    This is when it's really easy to do, and this is where you'll save the most time compared to using the quadratic formula or completing the square. Start by halving b, and voilà, you already have the real part of your complex roots (well, MINUS the real part)! Now subtract its square from c and square root the number you get, and that will be your imaginary part!

    Example:
    Let's do x^2+6x+13 together. We can easily see the real part is -3, and squaring that and taking it away from 13, 13-(-3)^2=13-9=4. So our imaginary part is \sqrt{4}=4, therefore our roots are -3\pm 2i .

    When a\neq 1 :
    There are a few ways to do this without the quadratic formula, but unless a is a perfect square and b is divisible by 2 AND \sqrt{a}, and that is a lot to ask, then these methods aren't worth it. So in the end I recommend using the quadratic formula. OF COURSE, here I'm assuming that a,b,\text{and } c have no factor in common.

    Try it!:
    (a,b,c)=
    (i)\ (1,4,5)
    (ii)\ (1,-5,8)
    (iii)\ (1,6,12)
    Solutions:
    Spoiler:
    Show
    (i)\ -2\pm i
    (ii)\ \frac{5\pm i \sqrt{7}}{2}
    (iii)\ -3\pm i\sqrt{3}
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    (Original post by gagafacea1)
    Example:
    Let's do x^2+6x+13 together. We can easily see the real part is -2, and squaring that and taking it away from 13, 13-(-2)^2=13-4=9. So our imaginary part is \sqrt{9}=3, therefore our roots are -2\pm 3i .
    When the person suggesting the method gets the wrong result, it's not a good sign.
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    (Original post by gagafacea1)
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    I don't really understand this post, you start off by saying that this method is helpful for people who aren't comfortable with the quadratic formula to factorise quadratic, then give a method that applies to a special case before ending with "use the quadratic formula".

    In any case, \displaystyle x^2 + 6x + 13 = 0 \iff x = -3 \pm 2i \neq -2 \pm 3i.

    I concur with ghostwalker on this one.
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    (Original post by ghostwalker)
    When the person suggesting the method gets the wrong result, it's not a good sign.
    Oh shoot, I'm so sorry. I knew this was gonna happen. Thanks for pointing it out anyways.
    (Original post by Zacken)
    I don't really understand this post, you start off by saying that this method is helpful for people who aren't comfortable with the quadratic formula to factorise quadratic, then give a method that applies to a special case before ending with "use the quadratic formula".

    In any case, \displaystyle x^2 + 6x + 13 = 0 \iff x = -3 \pm 2i \neq -2 \pm 3i.
    Tbh I was going to write down the method I use, but it might be a bit too much. Plus I wouldn't call a=1 a very special case. And really if you can't use the quadratic formula maybe you shouldn't be doing complex numbers just yet. What I meant was "didn't like the quadratic formula", so sorry about that.
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    Sorry OP but this is useless. I think vote downs should still be here for posts like this.
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    (Original post by zetamcfc)
    Sorry OP but this is useless. I think vote downs should still be here for posts like this.
    A bit too harsh to call this one of reasons for having vote downs. Many more actually useless posts out there.
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    To be honest I don't think this is a helpful post. There's only really one useful idea, (the a = 1 case), and this is really just a restatement of the quadratic formula taking advantage of the simplification that a = 1.

    But I would rather explicitly point out that the roots of x^2+2bx+c are -b \pm \sqrt{b^2-c} which holds for both real and complex roots.

    Situations where this applies are common enough that it's probably worth knowing about. (But it's obviously not essential, it just saves a small amount of calculation).
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    (Original post by DFranklin)
    To be honest I don't think this is a helpful post. There's only really one useful idea, (the a = 1 case), and this is really just a restatement of the quadratic formula taking advantage of the simplification that a = 1.

    But I would rather explicitly point out that the roots of x^2+2bx+c are -b \pm \sqrt{b^2-c} which holds for both real and complex roots.

    Situations where this applies are common enough that it's probably worth knowing about. (But it's obviously not essential, it just saves a small amount of calculation).
    I guess you're right, do you think I should delete the post? I only made it because a few people were surprised when I told them about it.
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    (Original post by gagafacea1)
    I guess you're right, do you think I should delete the post? I only made it because a few people were surprised when I told them about it.
    I don't think it's a big deal.

    I think it's fair to say that if you make a post giving tips / advice, then you are implicitly stating you're an authority, and those posts are going to get a lot more critical scrutiny, because we want to make sure you're not giving people bad advice (that they are likely to take as gospel).
 
 
 
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