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
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
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
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