The question in the image asks to find the value of $k$ such that the quadratic equation $x^2 + kx + 6 = 0$ has:

a) Two distinct roots
b) Double roots
c) No real roots

To solve this, we need to examine the discriminant $\Delta$ of the quadratic equation. The discriminant formula is given by:

$\Delta = b^2 - 4ac$

For the equation $ax^2 + bx + c = 0$, here $a = 1$, $b = k$, and $c = 6$. So, the discriminant becomes:

$\Delta = k^2 - 4(1)(6) = k^2 - 24$

Now, let's analyze the different cases based on the discriminant:

1. Two distinct roots (real and different): This occurs when $\Delta > 0$, meaning: $k^2 - 24 > 0$ $k^2 > 24$ Thus, $k > \sqrt{24}$ or $k < -\sqrt{24}$.
   This gives the values of $k$ as: $k > 2\sqrt{6} \quad \text{or} \quad k < -2\sqrt{6}$

2. Double roots (a repeated real root): This occurs when $\Delta = 0$, meaning: $k^2 - 24 = 0$ $k^2 = 24$ Thus, $k = \pm \sqrt{24}$.

3. No real roots: This occurs when $\Delta < 0$, meaning: $k^2 - 24 < 0$ $k^2 < 24$ Thus, $-\sqrt{24} < k < \sqrt{24}$.

Summary of conditions:

• For two distinct roots: $k > 2\sqrt{6}$ or $k < -2\sqrt{6}$.
• For double roots: $k = \pm \sqrt{24}$.
• For no real roots: $-\sqrt{24} < k < \sqrt{24}$.

Would you like further clarification or more questions about this topic?