Question 48

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waecmaths question: 

If ${{x}^{2}}+kx+\tfrac{16}{9}$ is a perfect square. Find the value of x

Option A: 

$\tfrac{8}{3}$

Option B: 

$\tfrac{7}{3}$

Option C: 

$\tfrac{5}{3}$

Option D: 

$\tfrac{2}{3}$

waecmaths solution: 

$\begin{align}  & {{x}^{2}}+kx+\tfrac{16}{9} \\ & \text{For perfect square to exist in a}{{\text{x}}^{2}}+bx+c=0 \\ & {{b}^{2}}=4ac \\ & a=1,b=k\text{ }c=\tfrac{16}{9} \\ & {{k}^{2}}=4(1)(\tfrac{16}{9}) \\ & {{k}^{2}}=\tfrac{64}{9} \\ & k=\pm \tfrac{8}{3} \\\end{align}$

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