Question

Find the discriminant of the equation and the nature of roots. Also find the roots.

$2x^2+5\sqrt{3}x+6=0$

• A. $D=27$, Real and distinct, $x=\frac{-\sqrt{3}}{2},-2\sqrt{3}$
• B. $D=0$, Real and equal, $x=\frac{-\sqrt{3}}{2}$
• C. $D=-2$, Imaginary, $x=\pm \frac{-\sqrt{3}}{2}i$
• D. Cannot be determined
  

Answer 1

The correct option is A $D=27$, Real and distinct, $x=\frac{-\sqrt{3}}{2},-2\sqrt{3}$

Dividing the entire equation by 2,, we get
$x^2+\frac{5\sqrt{3}x}{2}+3=0$
$(x+\frac{5\sqrt{3}}{4}{)}^2-\frac{75}{16}+\frac{48}{16}=0$
$(x+\frac{5\sqrt{3}}{4}{)}^2=\frac{27}{16}$
$x+\frac{5\sqrt{3}}{4}=\frac{\pm 3\sqrt{3}}{4}$
$x=\frac{-\sqrt{3}}{2}$ and $=-2\sqrt{3}$