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

1 Answer
Mar 26, 2017

Answer:

#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#