Answer: There are four complex zeros: `x=+-sqrt(3)i or x=+-i`
You can solve this a number of ways. The way I will show you is by using substitution and the quadratic formula.
Recall that the quadratic formula is based on the following quadratic formatted `ax^2+bx+c = 0`
Since we are looking for roots (when `f(x) = 0` ) then we are on the right track. However, this is a quartic not a quadratic! So, let's see if we can make it a quadratic using a substituion.
Let `u = x^2` Substitute `u` in for any instance of `x^2`
`f(x) = x^4+4x^2+3`
`f(u) = u^2+4u+3 `
Now we have a quadratic-looking equation where `a = 1` , `b=4` , and `c=3`
Using the quadratic formula:
`u=-3 or u=-1`
Now we have to substitute back
`x^2=-3 or x^2=-1`
Taking the squareroot of both sides, we end up with
since `i=sqrt(-1)` , then the answer becomes
This function has four complex roots, as the largest polynomial exponent suggests.
Just in case, here's a graph:
Given polynomial function f(x)=x^4+4x^2+3
let y be zero of f(x) i.e.
Thus roots are `+-i,+-isqrt(3)`