Question 2

We use the discriminant to find equal roots.

Step 1 — Write the equation

Let's look at the first equation. It is $2x^2 + kx + 3 = 0$.

Step 2 — Identify coefficients

We compare it to $ax^2 + bx + c = 0$. So, $\mathbf{a = 2}$, $\mathbf{b = k}$, and $\mathbf{c = 3}$.

Step 3 — Apply discriminant condition

For two equal roots, the discriminant must be zero. This means $b^2 - 4ac = 0$. Let's substitute the values.

$k^2 - 4(2)(3) = 0$

$k^2 - 24 = 0$

$k^2 = 24$

$k = \pm\sqrt{24}$

$k = \pm\sqrt{4 \times 6}$

$\boxed{k = \pm 2\sqrt{6}}$

Step 4 — Simplify the equation

Now, let's consider the second equation. It is $kx (x - 2) + 6 = 0$. We expand the expression.

$kx^2 - 2kx + 6 = 0$

Step 5 — Identify coefficients

We compare it to $ax^2 + bx + c = 0$. So, $\mathbf{a = k}$, $\mathbf{b = -2k}$, and $\mathbf{c = 6}$.

Step 6 — Apply discriminant condition

For two equal roots, the discriminant must be zero. This means $b^2 - 4ac = 0$. Let's substitute the values.

$(-2k)^2 - 4(k)(6) = 0$

$4k^2 - 24k = 0$

$4k(k - 6) = 0$

Step 7 — Solve for k

This gives us two possible values. Either $4k = 0$ or $k - 6 = 0$. So, $k = 0$ or $k = 6$.

Step 8 — Check for valid quadratic

If $\mathbf{k = 0}$, the original equation becomes $0 \cdot x(x - 2) + 6 = 0$. This simplifies to $6 = 0$. This is not a quadratic equation. So, $\mathbf{k}$ cannot be $\mathbf{0}$. Therefore, the only valid value for $\mathbf{k}$ is $\mathbf{6}$.

$\boxed{k = 6}$

Answer

(i) $k = \pm 2\sqrt{6}$ (ii) $k = 6$