Given that \(x^{2} + 4x + k = (x + r)^{2} + 1\), find the value of k and r.

Assessment: WAEC SSCE - Further Mathematics - 2015 (Objective) Subject: Further Mathematics

Question 1 Report

Given that \(x^{2} + 4x + k = (x + r)^{2} + 1\), find the value of k and r.

Answer Details
We know that the left side of the equation can be expanded to give: $$x^{2} + 4x + k = x^{2} + 2xr + r^{2} + 1$$ By comparing the coefficients of x and x^2, we get: $$2xr + 4x = 0 \implies x(2r+4) = 0$$ Since x cannot be equal to zero, we have: $$2r + 4 = 0 \implies r = -2$$ Substituting r = -2 in the original equation, we get: $$x^{2} + 4x + k = (x-2)^{2} + 1 = x^{2} - 4x + 5$$ Comparing the coefficients of x^2 and x, we get: $$k = 5$$ Therefore, the solution is k = 5, r = -2. Note that this matches with.
Read lesson note on Polynomial Functions (WAEC)
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