Question

This is my first time doing this. The question is in the photo.

Answer 1

SOLUTION:

Case: Roots of equation

Method:

$\begin{gathered} x^2-6x+12=0 \\ a=1,=-6,c=12 \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(12)}}{2(1)} \\ x=\frac{6\pm\sqrt{36-48}}{2} \\ x=\frac{6\pm\sqrt{-12}}{2} \\ x=\frac{6\pm\sqrt{-4\times3}}{2} \\ x=\frac{6\pm i2\sqrt{3}}{2} \\ x=3\pm i\sqrt{3} \end{gathered}$

Final answer: Option (D)

$x=3\pm\imaginaryI\sqrt{3}$