Given that \(x^{2} + 4x + k = (x + r)^{2} + 1\), find the value of k and r.
\(x^{2} + 4x + k = (x + r)^{2} + 1\)
\(x^{2} + 4x + k = x^{2} + 2rx + r^{2} + 1\)
Comparing the LHS and RHS equations, we have
\(2r = 4 \implies r = 2\)
\(k = r^{2} + 1 = 2^{2} + 1 = 5\)
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