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) \(kx (x - 2) + 6 = 0\)

(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.