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

$x = - \frac{1}{2} i$ and $x = + \frac{1}{2} i$ and $x = - 1$ and $x = + 1$

#### Explanation:

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

$\left(4 {x}^{2} + 1\right) = 0$ this is the first factor

$4 {x}^{2} = - 1 \text{ }$ divide both sides by 4

$\frac{\cancel{4} {x}^{2}}{\cancel{4}} = - \frac{1}{4} \text{ }$

${x}^{2} = - \frac{1}{4} \text{ }$take the square root of both sides

$\sqrt{{x}^{2}} = \pm \sqrt{- \frac{1}{4}}$

$x = \pm \frac{1}{2} i$

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For the other factor

$\left({x}^{2} - 1\right) = 0$

${x}^{2} = 1 \text{ " }$take the square root of both sides

$\sqrt{{x}^{2}} = \pm \sqrt{1}$

$x = \pm 1$

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

$x = - \frac{1}{2} i$ and $x = + \frac{1}{2} i$ and $x = - 1$ and $x = + 1$

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