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Solving Quadratics with imaginary roots

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3x^2 + 2x + 1 = 0 How would you solve this? When you put it into the quadratic formula, right away you can tell it's not going to work. If someone could also correct my errors in a completely naive attempt at solving this that would be awesome! I know I'm going wrong somewhere where it should be apparent that substituting isn't going to work, but I think something is fundamentally wrong about my approach here:

3x^2 + 2x + 1 = 0

let u = x ^ 2 and substitute

3u + 2x + 1 = 0
3u + 2x = -1
3u = -2x -1

u = -2x -1
    ------
       3

substitute u into the original equation

    -2x -1
3( --------- ) +2x + 1 = 0
       3

-2x -1 + 2x + 1 = 0

which leaves me with a badly designed proof for 

0 = 0

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Also, you can't substitute like that. All you're doing is essentially disregarding the original value and moving the rest of the stuff onto the right side, which leaves you with 0 = 0 as you noticed.

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If using the quadratic formula is out of the question you can always try completing the square(good google term). It's functionally equivalent, but might seem a little less magical and make more sense if you're new at this.

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I assumed it might be homework. Upon further inspection it would probably be more useful to say that when the quadratic formula "doesn't work" it actually does work and you'll just get imaginary solutions. When this happens it means that there aren't any real solutions.

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Quote:
Original post by Zipster
This smells a lot like homework, but I'll give you a hint. The determinant in the quadratic equation is going to be -8. √(-8) = √(8)√(-1). From there you can find the imaginary solutions.


It took me a while to realize why that sounded weird: The word is discriminant.

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I am learning this on my own time to fill gaps in my knowledge since I was taught math by my dad at "homeschool" - I'm not attending college or anything at the moment.

I don't have a very comprehensive background so that might be why it seems n00bish. Thanks for the help anyway. I have read the bits about imaginary numbers and am just not "getting" it - I was hoping someone could solve with a 'talk through' at each step.

The reason I say it's not going to 'work' in the quadratic equation is because of the sqrt on a negative; I am just not sure how to handle it. I understand i is sqrt(-1) - so from there, how do I represent sqrt(-2); surely it can't be simply 2i?

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okay, it's making a bit more sense now. x^2+3=0 means x = sqrt(-3)i; is there another way of notating it or is that the correct solution?


x^2 +3 = 0
x^2 = -3
x = sqrt(-3)i


Here's a simple one with an complex root I just tried...


x^2+2x+3 = 0
x^2+2x = - 3
x^2+2x+1 = -2
(x+1)^2 = -2
x+1 = sqrt(2)i
x = -1 + sqrt(2)i


That's as far as I should need to simplify?

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Quote:
Original post by djz
okay, it's making a bit more sense now. x^2+3=0 means x = sqrt(-3)i; is there another way of notating it or is that the correct solution?

No, it means x=sqrt(-3), which in turn is equal to sqrt(3)i.

Quote:
Here's a simple one with an complex root I just tried...


x^2+2x+3 = 0
x^2+2x = - 3
x^2+2x+1 = -2
(x+1)^2 = -2
x+1 = sqrt(2)i
x = -1 + sqrt(2)i


That's as far as I should need to simplify?

That one is correct, and there's nothing to simplify.

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You forgot the second solution of the quadratic equation. x2 - a = 0 => x2 = a => x = +- sqrt(a). In this case, x = +- sqrt(-3) = +- sqrt(3)i. When working with complex numbers, a quadratic equation always has two solutions which can coincide.

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The key is to realize that for a real number n, sqrt(-n) is equivalent to sqrt(n * i^2) because i^2 = -1 (and -1 * n = -n, obviously). sqrt(i^2) is i, leaving you with sqrt(n)*i.

Quote:
Original post by apatriarca
You forgot the second solution of the quadratic equation. x2 - a = 0 => x2 = a => x = +- sqrt(a). In this case, x = +- sqrt(-3) = +- sqrt(3)i. When working with complex numbers, a quadratic equation always has two solutions which can coincide.


To elaborate on that, if a polynomial function with real coefficients has one complex root a + bi, its conjugate (that is, a - bi) is also a root.

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