Answer:
The roots of the the given equation are +3i or -3i.
Solution:
Given, [tex]3x^2+27=0[/tex]
First of all compare the given equation with the standard form, i.e [tex]ax^2+bx+c=0[/tex]
On comparing,
a=3
b=0
c=27
According to the quadratic formula,
[tex]x = \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
[tex]b^2-4ac= 0-4\times3\times27=-324[/tex]
As [tex]b^2-4ac[/tex] comes out to be negative, the given equation does not have any real roots,
[tex]\sqrt {b^2-4ac}= \sqrt{(-324)}[/tex]
[tex]\sqrt {b^2-4ac}= \sqrt{(324\times -1)}[/tex]
[tex]\Rightarrow \sqrt{324}\times \sqrt{-1}[/tex]
Root of 324 is 18,
[tex]\Rightarrow 18\times i[/tex] (Because root of -1 is i)
[tex]x=\frac{(-0+-18i)}{2\times3}[/tex]
[tex]x=\frac{\pm18i}{6}[/tex]
[tex]x=\frac{+18}{6} \ or \frac{-18}{6}[/tex]
On dividing 18 and 6 we get,
[tex]\therefore x=+3i \ or -3i[/tex]
