 Q2. Find the value of k for each of the following quadratic equations, so that they have two equal roots.
(i) $$2x^2 + kx + 3 = 0$$
(ii) $$kx (x - 2) + 6 = 0$$

Ans.(i)$$2x^2 + kx + 3 = 0$$
We know that quadratic equation has two equal roots only when the value of discriminant is equal to zero.
Comparing equation with $$2x^2 + kx + 3 = 0$$ general quadratic equation $$ax^2 + bx + c = 0$$, we get a = 2, b = k and c = 3
Discriminant $$b^2 - 4ac = k^2 - 4 (2) (3) = k^2 - 24$$
Putting discriminant equal to zero
$$k^2 - 24= 0$$
$$k^2 = 24$$
$$k = ± \sqrt{2} = ± 2 \sqrt{6}$$

(ii)$$kx (x - 2) + 6 = 0$$
$$kx^2 - 2kx + 6 = 0$$
Comparing quadratic equation $$kx (x - 2) + 6 = 0$$ with general form $$ax^2 + bx + c = 0$$, we get a = k, b = ?2k and c = 6
Discriminant $$b^2 - 4ac = (-2k)^2 - 4 (k) (6) = 4k^2 - 24k$$
We know that two roots of quadratic equation are equal only if discriminant is equal to zero.
Putting discriminant equal to zero
$$4k^2 - 24k = 0$$
$$4k (k - 6) = 0=> k = 0, 6$$
The basic definition of quadratic equation says that quadratic equation is the equation of the form $$ax^2 + bx + c = 0$$, where a ? 0.
Therefore, in equation , we cannot have k = 0.
Therefore, we discard k = 0.
Hence the answer is k = 6.