# How do you solve x^4 - 3x^2 + 4 = 0 using the quadratic formula?

Aug 2, 2017

The solutions are $S = \left\{\pm \sqrt{\frac{3 \pm i \sqrt{7}}{2}}\right\}$

#### Explanation:

Our equation is

${x}^{4} - 3 {x}^{2} + 4 = 0$

This is a biquadratic equation

Let $X = {x}^{2}$

Then,

${X}^{2} - 3 X + 4 = 0$

We solve this like quadratic equation

The discriminant is $\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - 4 \left(1\right) \left(4\right) = - 7$

As $\Delta < 0$, the solutions are in $\mathbb{C}$

$X = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{3 \pm i \sqrt{7}}{2}$

Therefore,

$x = \pm \sqrt{X} = \pm \sqrt{\frac{3 \pm i \sqrt{7}}{2}}$

Aug 2, 2017

Substitute $p$ for ${x}^{2}$, so $p = {x}^{2}$. The quadratic equation we get is then:

${p}^{2} - 3 p + 4 = 0$

Note that this equation has complex roots. Then, using these roots (${p}_{1}$ and ${p}_{2}$), calculate the values of $x$ using the fact that $p = {x}^{2}$.

so:

${x}_{1} = \sqrt{{p}_{1}}$
${x}_{2} = - \sqrt{{p}_{1}}$
${x}_{3} = \sqrt{{p}_{2}}$
${x}_{4} = - \sqrt{{p}_{2}}$

Aug 2, 2017

$x = \pm \sqrt{\frac{3 \pm \sqrt{5} i}{2}}$

#### Explanation:

Let $r = {x}^{2}$

${x}^{4} - 3 {x}^{2} + 4 = 0 \textcolor{w h i t e}{\text{xx")harrcolor(white)("xxx}} {r}^{2} - 3 r + 4 = 0$

We can now apply the quadratic formula to the equation in term of $r$
$\textcolor{w h i t e}{\text{XXX}} r = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot \left(1\right) \cdot \left(4\right)}}{2 \cdot \left(1\right)}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{3 \pm \sqrt{- 5}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{3 \pm \sqrt{5} i}{2}$

and since $r = {x}^{2} \textcolor{w h i t e}{\text{xx")rarrcolor(white)("xx}} x = \pm \sqrt{r}$
we have
$\textcolor{w h i t e}{\text{XXX}} x = \pm \sqrt{\frac{3 \pm \sqrt{5} i}{2}}$