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Math Help - Real Roots of a Polynomial Equation

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    Real Roots of a Polynomial Equation

    How do you find a real root of an equation? I thought when you get the roots for example y=(x-3)(x+2)(x-7) the roots would be 3,-2, and 7. I'm not sure of how to know if it is "real" or not.
    How would you go about finding the real root of (5x^2+20)(3x^2-48)=0
    and (2x^2-x-13)(x^2+1)= 0
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    Quote Originally Posted by kimvo76 View Post
    How do you find a real root of an equation? I thought when you get the roots for example y=(x-3)(x+2)(x-7) the roots would be 3,-2, and 7. I'm not sure of how to know if it is "real" or not.
    How would you go about finding the real root of (5x^2+20)(3x^2-48)=0
    and (2x^2-x-13)(x^2+1)= 0
    see the link for an explanation of imaginary numbers (numbers that are not real) ...

    Complex Numbers: Introduction



    (5x^2+20)(3x^2-48)=0

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

    roots are ...

    x = \pm 2i , x = \pm 4

    which roots are real?


    (2x^2-x-13)(x^2+1) = 0

    (2x^2-x-13)(x+i)(x-i)= 0

    \displaystyle x = \frac{1 \pm \sqrt{105}}{4} , x = \pm i

    which roots are real?
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