Question

How do you find all the real and complex roots of `-4x^4+3x^2+1`?

Answer 1

`x=-1/2i` and `x=+1/2i` and `x=-1` and `x=+1`

Explanation:

Let us start from the given
`-4x^4+3x^2+1=0`
this is also
`4x^4-3x^2-1=0`
Using Factoring Method
`(4x^2+1)(x^2-1)=0`
and equating both factors to zero we have

`(4x^2+1)=0` this is the first factor

`4x^2=-1" "` divide both sides by 4

`(cancel4x^2)/cancel4=-1/4" "`

`x^2=-1/4" "`take the square root of both sides

`sqrt(x^2)=+-sqrt(-1/4)`

`x=+-1/2i`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For the other factor

`(x^2-1)=0`

`x^2=1" " "`take the square root of both sides

`sqrt(x^2)=+-sqrt(1)`

`x=+-1`

Just because this is a 4th degree equation, we are expecting 4 possible roots or zeros

`x=-1/2i` and `x=+1/2i` and `x=-1` and `x=+1`

God bless...I hope the explanation is useful..