Question

If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of \(k\) is

• A. \(-2\)
• B. \(\sqrt{\frac{8}{3}}\)
• C. \(1\)
• D. \(2\)

Hint:

When a quadratic has "equal roots", it's a perfect square trinomial. The middle coefficient squared always equals \(4 \times \text{first term} \times \text{last term}\).

Answer 1

The Correct Option is B

To determine the value of \( k \) such that the roots of the quadratic equation \( x^2 - k\sqrt{3}x + 2 = 0 \) are real and equal, we must use the condition for equal roots in a quadratic equation. The condition is that the discriminant (\( D \)) should be equal to zero.

The general form of a quadratic equation is:

\(ax^2 + bx + c = 0\)

where \( a = 1 \), \( b = -k\sqrt{3} \), and \( c = 2 \).

The discriminant \( D \) is given by:

\(D = b^2 - 4ac\)

Substituting the values from the given quadratic equation:

• \( a = 1 \)
• \( b = -k\sqrt{3} \)
• \( c = 2 \)

The discriminant becomes:

\(D = (-k\sqrt{3})^2 - 4 \times 1 \times 2\)

Simplifying it further:

\(D = k^2 \times 3 - 8\)

For the roots to be real and equal, the discriminant must be zero:

\(k^2 \times 3 - 8 = 0\)

Simplifying the equation:

\(3k^2 = 8\)

Solving for \( k^2 \):

\(k^2 = \frac{8}{3}\)

Taking the square root on both sides:

\(k = \pm\sqrt{\frac{8}{3}}\)

Since the options provided only contain positive values, the solution is:

\(k = \sqrt{\frac{8}{3}}\)

Thus, the correct answer is:

\(\sqrt{\frac{8}{3}}\)