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) \(kx(x - 2) + 6 = 0\)
(i) \(k = \pm 2\sqrt{6}\)
(ii) \(k = 6\)
Key fact: For a quadratic equation \(ax^2 + bx + c = 0\), the roots are equal (a repeated root) exactly when the discriminant is zero.
Discriminant: \(D = b^2 - 4ac\).
Condition for equal roots: \(D = 0\).
Step 1: Identify \(a, b, c\).
Compare with \(ax^2 + bx + c = 0\):
\[a = 2, \quad b = k, \quad c = 3\]
Step 2: Write the discriminant.
\[D = b^2 - 4ac = k^2 - 4(2)(3)\]
Simplify:
\[D = k^2 - 24\]
Step 3: Use the equal-root condition.
For equal roots, set \(D = 0\):
\[k^2 - 24 = 0\]
\[k^2 = 24\]
Step 4: Solve for \(k\).
\[k = \pm \sqrt{24} = \pm \sqrt{4 \cdot 6} = \pm 2\sqrt{6}\]
Conclusion for (i): \(k = \pm 2\sqrt{6}\).
Step 1: Expand to standard quadratic form.
First expand \(kx(x - 2)\):
\[kx(x - 2) = kx^2 - 2kx\]
So the equation becomes:
\[kx^2 - 2kx + 6 = 0\]
Step 2: Identify \(a, b, c\).
Comparing with \(ax^2 + bx + c = 0\):
\[a = k, \quad b = -2k, \quad c = 6\]
Step 3: Write the discriminant.
\[D = b^2 - 4ac = (-2k)^2 - 4(k)(6)\]
Simplify each term:
\[(-2k)^2 = 4k^2\]
\[4(k)(6) = 24k\]
So,
\[D = 4k^2 - 24k\]
Step 4: Use the equal-root condition.
Set \(D = 0\):
\[4k^2 - 24k = 0\]
Factor out \(4k\):
\[4k(k - 6) = 0\]
Step 5: Solve for \(k\).
From \(4k = 0\), we get \(k = 0\).
From \(k - 6 = 0\), we get \(k = 6\).
Step 6: Check the validity of \(a \neq 0\). For a quadratic equation, \(a\) must not be zero. Here \(a = k\). If \(k = 0\), the equation becomes:
\[0 \cdot x^2 - 0 \cdot x + 6 = 0 \Rightarrow 6 = 0\]
which is impossible. So \(k = 0\) is not allowed.
Therefore, the only acceptable value is:
\[k = 6\]
Conclusion for (ii): \(k = 6\) gives equal roots.