3 Tutor System
Starting just at 265/hour

# 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

$$\Rightarrow k^2 - 24= 0$$
$$\Rightarrow k^2 = 24$$
$$\Rightarrow k = ± \sqrt{2} = ± 2 \sqrt{6}$$

(ii)$$kx (x - 2) + 6 = 0$$
$$\Rightarrow kx^2 - 2kx + 6 = 0$$

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
$$\Rightarrow 4k^2 - 24k = 0$$
$$\Rightarrow 4k (k - 6) = 0$$
$$\Rightarrow k = 0, 6$$

The basic definition of quadratic equation says that quadratic equation is the equation of the form $$ax^2 + bx + c = 0$$, where a $$\neq$$ 0.

Therefore, in equation , we cannot have k = 0.
Therefore, we discard k = 0.
Hence the answer is k = 6.