Real Roots of a Polynomial Equation

Quote:

Originally Posted by kimvo76

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?