Question

How to solve X^4-19x^2+48=0

Answer 1

Two numbers that add up to -19 and multiply to 48 are -16 and -3:

$(x^4-19x^2+48)=0\\(x^2-16)(x^2-3)=0\\(x+4)(x-4)(x^2-3)=0$

So, the solutions come from each parentheses:  x+4=0, x-4=0, and x^2-3=0.

x+4=0
x = -4

x-4=0
x = 4

x^2-3=0
x^2 = 3
x = +/- √3

So, the solutions are -4, -√3, √3, and 4.