Question

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

Answer 1

`x = (-3+sqrt(3)+-i (sqrt(3)-1))/2`

Explanation:

`(2+sqrt(3))x^2+(3+sqrt(3))x+2 = 0`

is in the standard form:

`ax^2+bx+c = 0`

with `a=2+sqrt(3)`, `b = 3+sqrt(3)` and `c=2`

Let's look at the discriminant first...

`Delta = b^2-4ac`

`color(white)(Delta) = (3+sqrt(3))^2-4(2+sqrt(3))(2)`

`color(white)(Delta) = 9+6sqrt(3)+3-16-8sqrt(3)`

`color(white)(Delta) = -4-2sqrt(3)`

Since `Delta < 0` this quadratic has no real solutions.

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

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (-b+-sqrt(Delta))/(2a)`

`color(white)(x) = (-3-sqrt(3)+-sqrt(-4-2sqrt(3)))/(4+2sqrt(3))`

`color(white)(x) = (-3-sqrt(3)+-i sqrt(4+2sqrt(3)))/(4+2sqrt(3))`

`color(white)(x) = ((-3-sqrt(3)+-i sqrt(4+2sqrt(3)))(2-sqrt(3)))/((4+2sqrt(3))(2-sqrt(3)))`

`color(white)(x) = ((-3-sqrt(3)+-i sqrt(2(2+sqrt(3))))(2-sqrt(3)))/2`

`color(white)(x) = (-3+sqrt(3)+-i sqrt(2(2-sqrt(3))))/2`

Note that:

`(sqrt(3)-1)^2 = sqrt(3)^2-2sqrt(3)+1 = 4-2sqrt(3)`

So:

`sqrt(2(2-sqrt(3))) = sqrt(3)-1`

So:

`x = (-3+sqrt(3)+-i (sqrt(3)-1))/2`