Question

Solve $\sqrt{3}x^2 – \sqrt{2} x + 3\sqrt{3}$ = 0

(a) $\frac{-1±i\sqrt{17}}{2}$

(b) $\frac{1±i\sqrt{17}}{2}$

(c) $\frac{1±\sqrt{17}}{2}$

(d) $\frac{-1±\sqrt{17}}{2}$

I got this question during an interview for a job.

The question is from Quadratic Equations topic in section Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer 1

Right choice is (b) $\frac{1±i\sqrt{17}}{2}$

The explanation is: $\sqrt{3}x^2 – \sqrt{2} x + 3\sqrt{3}$ = 0

=>3x^2 – $\sqrt{6}$x + 9 = 0

=>D=(-$\sqrt{6}$)^2 – 4.3.9 = 6-108 = -102.

Since D ≤ 0, imaginary roots are there.

=>x = $\frac{\sqrt{6}±i\sqrt{102}}{2.3} = \frac{1±i\sqrt{17}}{2}$.