Chapter: Quadratic Equations· Mathematics Part 1 (Algebra)
Answer: \(k = 5\) (root \(-\dfrac{1}{3}\)) or \(k = -3\) (root 1)
\(a = k + 4\), \(b = k + 1\), \(c = 1\). Equal roots \(\Rightarrow b^2 - 4ac = 0\).
\((k + 1)^2 - 4(k + 4) = 0\)
\(k^2 + 2k + 1 - 4k - 16 = 0 \Rightarrow k^2 - 2k - 15 = 0\)
\((k - 5)(k + 3) = 0 \Rightarrow k = 5\) or \(k = -3\)
Neither value makes \(k + 4 = 0\), so both are valid.
For \(k = 5\): \(9x^2 + 6x + 1 = 0 \Rightarrow (3x + 1)^2 = 0 \Rightarrow x = -\dfrac{1}{3}\).
For \(k = -3\): \(x^2 - 2x + 1 = 0 \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1\).