Find the value of $k$ for which the quadratic equation $k x(x - 2) + 6 = 0$ has two equal roots.
Step-by-Step Solution
Key Concept: Rewrite as $kx^2 - 2kx + 6 = 0$. For equal roots, $D = b^2 - 4ac = 0$. Note $k <br>eq 0$ for a quadratic.
Stepwise Solution:
Standard form: $kx^2 - 2kx + 6 = 0$. Here $a=k, b=-2k, c=6$. [0.5 Mark]
$D = (-2k)^2 - 4(k)(6) = 4k^2 - 24k$. [0.5 Mark]
For equal roots, $D = 0 \Rightarrow 4k(k - 6) = 0 \Rightarrow k = 0$ or $k = 6$.
Since $k=0$ makes the coefficient of $x^2$ zero (not a quadratic), we reject $k=0$. Thus $k = 6$. [1.0 Mark]
Marking Scheme:
• Standard form and identifying coefficients: 0.5 Mark
• Forming discriminant equation $4k^2 - 24k = 0$: 0.5 Mark
• Solving for $k = 6$ and rejecting $k = 0$: 1.0 Mark
Correct Answer: