How do you solve (2+sqrt(3))x^2+(3+sqrt(3))x+2 = 0(2+3)x2+(3+3)x+2=0 ?

1 Answer
Mar 26, 2017

x = (-3+sqrt(3)+-i (sqrt(3)-1))/2x=3+3±i(31)2

Explanation:

(2+sqrt(3))x^2+(3+sqrt(3))x+2 = 0(2+3)x2+(3+3)x+2=0

is in the standard form:

ax^2+bx+c = 0ax2+bx+c=0

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