Find the discriminant of the equation and the nature of roots. Also find the roots. $2x^2 - 6x + 3 = 0$
- $D=12$, Roots are real and distinct, $x=\displaystyle \frac{3 + \sqrt 3}{2}, \frac{3 - \sqrt 3}{2}$
- $D=0$, Roots are real and equal, $x=\displaystyle \frac{3 + \sqrt 3}{2}$
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$D=-9$, Roots are imaginary.
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None of these
For 2x^2 - 6x + 3 = 0, the discriminant D = b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 - 24 = 12. Since D > 0, the roots are real and distinct, calculated via the quadratic formula as (6 +/- sqrt(12))/4 = (3 +/- sqrt(3))/2.
The discriminant formula is D equals b squared minus 4ac. Substituting the coefficients a equals 2, b equals negative 6, and c equals 3 gives D equals 36 minus 24, which is 12. Since the discriminant is positive, the roots are real and distinct. Using the quadratic formula, the roots evaluate to x equals the quantity (3 plus the square root of 3) divided by 2 and x equals the quantity (3 minus the square root of 3) divided by 2.