WAEC - Further Mathematics (2015 - No. 19)
Given that \(x^{2} + 4x + k = (x + r)^{2} + 1\), find the value of k and r.
k = 5, r = -1
k = 5, r = 2
k = 2, r = -5
k = -1, r = 5
Selitys
\(x^{2} + 4x + k = (x + r)^{2} + 1\)
\(x^{2} + 4x + k = x^{2} + 2rx + r^{2} + 1\)
Comparing the LHS and RHS equations, we have
\(2r = 4 \implies r = 2\)
\(k = r^{2} + 1 = 2^{2} + 1 = 5\)