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