How to determine real roots of (5x^2+20)(3x^2-48)=0?

Question

How to determine real roots of (5x^2+20)(3x^2-48)=0?

Answer 1

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

 

Use zero-product rule:

 

Case 1: 5x^2+20 = 0; 5x^2 = -20; x^2 = -4; |x| = 2*sqrt(-1); x = +-2i

Case 2: 3x^2-48 = 0; 3x^2 = 48; x^2 = 16; |x| = 4; x = +-4

 

Real Roots are x = +-4 and the Imaginary Roots are x = +-2i.