Question

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

Answer 1

Case: Roots of equation

Method:

`\begin{gathered} x^2-6x+12=0 \\ a=1,=-6,c=12 \\ x=(-b\pm√(b^2-4ac))/(2a) \\ x=(-(-6)\pm√((-6)^2-4(1)(12)))/(2(1)) \\ x=(6\pm√(36-48))/(2) \\ x=(6\pm√(-12))/(2) \\ x=(6\pm√(-4*3))/(2) \\ x=(6\pm i2√(3))/(2) \\ x=3\pm i√(3) \end{gathered}`

Final answer: Option (D)

`x=3\pm\imaginaryI√(3)`