# How do you solve (2+sqrt(3))x^2+(3+sqrt(3))x+2 = 0 ?

Mar 26, 2017

$x = \frac{- 3 + \sqrt{3} \pm i \left(\sqrt{3} - 1\right)}{2}$

#### Explanation:

$\left(2 + \sqrt{3}\right) {x}^{2} + \left(3 + \sqrt{3}\right) x + 2 = 0$

is in the standard form:

$a {x}^{2} + b x + c = 0$

with $a = 2 + \sqrt{3}$, $b = 3 + \sqrt{3}$ and $c = 2$

Let's look at the discriminant first...

$\Delta = {b}^{2} - 4 a c$

$\textcolor{w h i t e}{\Delta} = {\left(3 + \sqrt{3}\right)}^{2} - 4 \left(2 + \sqrt{3}\right) \left(2\right)$

$\textcolor{w h i t e}{\Delta} = 9 + 6 \sqrt{3} + 3 - 16 - 8 \sqrt{3}$

$\textcolor{w h i t e}{\Delta} = - 4 - 2 \sqrt{3}$

Since $\Delta < 0$ this quadratic has no real solutions.

We can still use the quadratic formula to find the complex solutions:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- 3 - \sqrt{3} \pm \sqrt{- 4 - 2 \sqrt{3}}}{4 + 2 \sqrt{3}}$

$\textcolor{w h i t e}{x} = \frac{- 3 - \sqrt{3} \pm i \sqrt{4 + 2 \sqrt{3}}}{4 + 2 \sqrt{3}}$

$\textcolor{w h i t e}{x} = \frac{\left(- 3 - \sqrt{3} \pm i \sqrt{4 + 2 \sqrt{3}}\right) \left(2 - \sqrt{3}\right)}{\left(4 + 2 \sqrt{3}\right) \left(2 - \sqrt{3}\right)}$

$\textcolor{w h i t e}{x} = \frac{\left(- 3 - \sqrt{3} \pm i \sqrt{2 \left(2 + \sqrt{3}\right)}\right) \left(2 - \sqrt{3}\right)}{2}$

$\textcolor{w h i t e}{x} = \frac{- 3 + \sqrt{3} \pm i \sqrt{2 \left(2 - \sqrt{3}\right)}}{2}$

Note that:

${\left(\sqrt{3} - 1\right)}^{2} = {\sqrt{3}}^{2} - 2 \sqrt{3} + 1 = 4 - 2 \sqrt{3}$

So:

$\sqrt{2 \left(2 - \sqrt{3}\right)} = \sqrt{3} - 1$

So:

$x = \frac{- 3 + \sqrt{3} \pm i \left(\sqrt{3} - 1\right)}{2}$