waecmaths question:
Given that one of the roots of the equation $2{{x}^{2}}+(k+2)x+k=0$ is 2. Find the value of k
Option A:
–4
Option B:
–2
Option C:
–1
Option D:
$-\tfrac{1}{4}$
waecmaths solution:
$\begin{align} & \text{Let }f(x)=2{{x}^{2}}+(k+2)x+k \\ & \text{Since }2\text{ is a root of the equation then }f(-2)=0 \\ & f(-2)=2{{(-2)}^{2}}+(k+2)(-2)+k=0 \\ & 8+(-2k-4)+k=0 \\ & 4-k=0 \\ & k=4 \\\end{align}$
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