Question

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

• A. $\pm \sqrt{24}$
• B. 0
• C. 4
• D. -5

Hint:

For equal roots, $b^2 = 4ac$.
You can quickly calculate $4ac$ in your head: $4 \times \sqrt{3} \times 2\sqrt{3} = 4 \times 6 = 24$.
So, $k^2 = 24 \implies k = \pm\sqrt{24}$.
Be careful not to forget the $\pm$ sign, as a quadratic equation with a squared term generally yields two possible values for its parameters!

Answer 1

The Correct Option is A

Step 1: Normalize the equation first by dividing through by $\sqrt{3}$, instead of working with the original coefficients directly.
The given equation is:
\[ \sqrt{3}x^2 - kx + 2\sqrt{3} = 0 \]
Divide every term by $\sqrt{3}$ to make the leading coefficient equal to $1$:
\[ x^2 - \frac{k}{\sqrt{3}}x + 2 = 0 \]

Step 2: Identify the new coefficients.
Comparing with $ax^2+bx+c=0$:
\[ a = 1, \qquad b = -\frac{k}{\sqrt{3}}, \qquad c = 2 \]

Step 3: Apply the condition for real and equal roots on these simplified coefficients.
For real and equal roots, the discriminant must be zero:
\[ b^2 - 4ac = 0 \]
\[ \left(-\frac{k}{\sqrt{3}}\right)^2 - 4(1)(2) = 0 \]
\[ \frac{k^2}{3} - 8 = 0 \]

Step 4: Solve for k.
\[ \frac{k^2}{3} = 8 \]
\[ k^2 = 24 \]
\[ k = \pm\sqrt{24} \]

Final Answer:
The values of $k$ for which the roots are real and equal are $\pm\sqrt{24}$, which matches Option (A). \[ \boxed{k = \pm\sqrt{24}} \]