# Real Roots of a Polynomial Equation

• Oct 5th 2010, 04:28 PM
kimvo76
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
• Oct 5th 2010, 04:58 PM
skeeter
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

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

$\displaystyle 15(x+2i)(x-2i)(x+4)(x-4) = 0$

roots are ...

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

which roots are real?

$\displaystyle (2x^2-x-13)(x^2+1) = 0$

$\displaystyle (2x^2-x-13)(x+i)(x-i)= 0$

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

which roots are real?