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

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.