2X^2=56,,,,,,,,,,#2 is find all real/complex roots for x^4-4x^2+3=0?

Question

PLEASE HELP ASAP

Answer 1

x^4 - 4x^2 + 3 = 0

First, treat it like a quadratic; note that x^4 can be decomposed as
x^2 and x^2. Factor it as such.

(x^2 - 3)(x^2 - 1) = 0

Now, equate each of these factors to 0 and solve individually.

x^2 - 3 = 0
x^2 = 3
x = +/- sqrt(3) {Note: Whenever you take the square root of both sides of an equation, always include a "plus or minus"}

x^2 - 1 = 0
(x - 1) (x + 1) = 0, which implies x = {-1, 1}

Therefore, our four solutions are

x = {sqrt(3),-sqrt(3), 1, -1}