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Math Help - quadratics and imaginary roots

  1. #1
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    quadratics and imaginary roots

    Hi;
    If a quadratic polynomial won't factor does this mean it
    has imaginary roots?

    if I complete the square on the same polynomial which makes it
    factorable does this process give it real roots?
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  2. #2
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    Re: quadratics and imaginary roots

    Quote Originally Posted by anthonye View Post
    Hi;
    If a quadratic polynomial won't factor does this mean it
    has imaginary roots?
    if I complete the square on the same polynomial which makes it
    factorable does this process give it real roots?
    Consider 3x^2-3x-2, is that what you mean?
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  3. #3
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    Re: quadratics and imaginary roots

    Yes
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    Re: quadratics and imaginary roots

    Quote Originally Posted by anthonye View Post
    Hi;
    If a quadratic polynomial won't factor does this mean it
    has imaginary roots?
    If it won't factor over the reals, then yes, it will have COMPLEX roots (which have a real part and an imaginary part).

    if I complete the square on the same polynomial which makes it
    factorable does this process give it real roots?
    Completing the square won't magically make a quadratic factorise over the reals. You WILL however be able to factorise over the complex numbers, which will still give complex roots.
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    Re: quadratics and imaginary roots

    Ok thank you Plato and prove it.
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    Re: quadratics and imaginary roots

    Quote Originally Posted by anthonye View Post
    Yes
    The point is that any quadratic with real coefficients can be factored.
    Take the example I gave you. 3x^2-3x-2=0 has two real roots: \left( {\frac{{3 + \sqrt {33} }}{6}} \right)\;\& \;\left( {\frac{{3 - \sqrt {33} }}{6}} \right).

    The factored form is: 3x^2-3x-2=\left[ {x - \left( {\frac{{3 + \sqrt {33} }}{6}} \right)} \right]\left[ {x - \left( {\frac{{3 - \sqrt {33} }}{6}} \right)} \right]=0.
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    Re: quadratics and imaginary roots

    On the other hand, x^2+1 has no real zeros but also can be factored: (x- i)(x+ i). The trouble is, you didn't say what you meant by "factored".
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    Re: quadratics and imaginary roots

    Please can you elaborate on the fact that any quadratic with real coefficients can be factored

    I'm a little lost with that.
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    Re: quadratics and imaginary roots

    Any quadratic, whether its coefficients are real or not, can be factored- but NOT necessarily with integer or even real coefficients.

    Take ax^2+ bx+ c as a generic example. First, factor out the "a". a(x^2+ \frac{b}{a}x+ \frac{c}{a}). Now "complete the square"- add and subtract \frac{b^2}{4a^2}: a(x^2+ \frac{b}{a}x+ \frac{b^2}{4a}+ \frac{c}{a}- \frac{b^2}{4a^2}). The point of that is that x^2+ \frac{b}{a}x+ \frac{b^2}{4a^2}= (x+ \frac{b}{2a})^2, a perfect square. And of course, \frac{b^2}{4ac^2}-\frac{c}{a}= \left(\sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)^2.

    That is, we can write ax^2+ bx+ c= a\left(\left(x+ \frac{b}{2a}\right)^2- \left(\sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)^2\right)

    And now we use the fact that x^2- a^2= (x- a)(x+ a) to write
    a\left(x+ \frac{b}{2a}+\sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)\left(x+ \frac{b}{2a}- \sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)

    That is factored into a number times two linear factors. With those square roots, those numbers may not be integers or even real numbers.
    Last edited by HallsofIvy; August 15th 2012 at 05:31 AM.
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    Re: quadratics and imaginary roots

    Wow ok thanks Hallsofivy Ineed to look at this more any idea where I can read about this subject?
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