The roots of the quadratic equation $2x^2 + kx + 3$ = 0 are equal, then value of 'k' is:

Question

Accepted Answer

To determine the value of 'k' for which the roots of the quadratic equation $2x^2 + kx + 3 = 0$ are equal, we need to use the condition for equal roots in a quadratic equation. The roots of a quadratic equation $ax^2 + bx + c = 0$ are equal if and only if the discriminant is zero. The discriminant $\Delta$ is given by:

$$
\Delta = b^2 - 4ac
$$

For the given equation $2x^2 + kx + 3 = 0$, we have $a = 2$, $b = k$, and $c = 3$. Substituting these values into the formula for the discriminant, we get:

$$
\Delta = k^2 - 4 \times 2 \times 3 = k^2 - 24
$$

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

$$
k^2 - 24 = 0
$$

Solving for $k$, we add 24 to both sides:

$$
k^2 = 24
$$

Taking the square root of both sides, we find:

$$
k = \pm \sqrt{24}
$$

Simplifying $\sqrt{24}$, we have:

$$
\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}
$$

Thus, the value of $k$ is $\pm 2\sqrt{6}$.

Now, let's justify why Option 1 is correct and why the other options are incorrect:

1. **Option 1: $\pm 2\sqrt{6}$** - This is the correct answer because, as derived above, the value of $k$ that makes the discriminant zero is indeed $\pm 2\sqrt{6}$.

2. **Option 2: $\pm 3\sqrt{8}$** - This option is incorrect because $\sqrt{8}$ simplifies to $2\sqrt{2}$, making $3\sqrt{8} = 6\sqrt{2}$, which does not satisfy the equation $k^2 = 24$.

3. **Option 3: $\pm 5\sqrt{6}$** - This option is incorrect because if $k = 5\sqrt{6}$, then $k^2 = 25 \times 6 = 150$, which is not equal to 24.

4. **Option 4: $\pm 4\sqrt{3}$** - This option is incorrect because if $k = 4\sqrt{3}$, then $k^2 = 16 \times 3 = 48$, which is not equal to 24.

Therefore, the only value of $k$ that satisfies the condition for equal roots is $\pm 2\sqrt{6}$, making Option 1 the correct choice.