Question 26

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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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