Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) \(2x^2 + kx + 3 = 0\) (ii) \(kx (x – 2) + 6 = 0\)
A quadratic equation \(ax^2 + bx + c = 0\) has two equal roots if its discriminant \((b^2 − 4ac)\) is equal to 0.
(i) For the equation \(2x^2 + kx + 3 = 0\):
Comparing with \(ax^2 + bx + c = 0\), we have a = 2, b = k, and c = 3.
The discriminant is \(b^2 − 4ac = (k)^2− 4(2)(3) = k^2 − 24\).
For equal roots, the discriminant must be 0:
\(k^2 − 24 = 0\)
\(k^2 = 24\)
\(k = ±\sqrt{24} = ±2\sqrt6\)
(ii) For the equation \(kx (x − 2) + 6 = 0\), which simplifies to \(kx^2 − 2kx + 6 = 0\):
Comparing with \(ax^2 + bx + c = 0\), we have a = k, b = −2k, and c = 6.
The discriminant is \(b^2 − 4ac = (− 2k)^2 − 4(k)(6) = 4k^2 − 24k\).
For equal roots, the discriminant must be 0:
\(4k^2 − 24k = 0\)
\(4k(k − 6) = 0\)
This implies either \(4k = 0\) or \(k − 6 = 0\), leading to \(k = 0\) or \(k = 6\).
If \(k = 0\), the equation loses its \(x^2\) and \(x\) terms, thus not being a quadratic equation.
Therefore, for this equation to have two equal roots, \(k\) must be 6.