Find the discriminant of the equation and the nature of roots. Also find the roots. $2x^2 + 5 \sqrt 3x + 6 =0$
- $D=27$, Real and distinct, $x=\displaystyle \frac{-\sqrt 3}{2}, -2 \sqrt 3$
- $D=0$, Real and equal, $x=\displaystyle \frac{-\sqrt 3}{2}$
- $D=-2$, Imaginary, $x= \pm \displaystyle \frac{-\sqrt 3}{2}i$
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Cannot be determined
Equation: 2x^2 + 5*sqrt(3)x + 6 = 0. Discriminant D = b^2 - 4ac = (5*sqrt(3))^2 - 4(2)(6) = 75 - 48 = 27. Since D > 0, roots are real and distinct. Roots = (-b +/- sqrt(D)) / 2a = (-5*sqrt(3) +/- sqrt(27)) / 4 = (-5*sqrt(3) +/- 3*sqrt(3)) / 4. Roots are -2*sqrt(3)/4 = -sqrt(3)/2 and -8*sqrt(3)/4 = -2*sqrt(3).
The discriminant D is calculated as b squared minus 4ac, which is 75 minus 48, giving D equal to 27. Because the discriminant is positive, the equation has real and distinct roots. Using the quadratic formula, the roots are negative 5 times the square root of 3 plus or minus 3 times the square root of 3, all divided by 4, which simplifies to negative 2 times the square root of 3 and negative the square root of 3 divided by 2.