Question

Find the roots of the quadratic equation \(3x^2 - 2\sqrt{6}x + 2 = 0\).

Hint:

When the discriminant \(D = 0\), both roots are real and equal.

Answer 1

Step 1: Identify coefficients.
\[ a = 3, \; b = -2\sqrt{6}, \; c = 2 \]
Step 2: Apply the quadratic formula.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Step 3: Substitute the values.
\[ x = \frac{2\sqrt{6} \pm \sqrt{(-2\sqrt{6})^2 - 4(3)(2)}}{2(3)} \] \[ x = \frac{2\sqrt{6} \pm \sqrt{24 - 24}}{6} \] \[ x = \frac{2\sqrt{6} \pm 0}{6} = \frac{\sqrt{6}}{3} \]
Step 4: Conclusion.
Hence, both roots are equal and given by \[ \boxed{x = \frac{\sqrt{6}}{3}} \]