waecmaths question:
What must be added to ${{x}^{2}}-3x$ to make it a perfect square?
Option A:
$\tfrac{9}{4}$
Option B:
$\tfrac{9}{2}$
Option C:
6
Option D:
9
waecmaths solution:
$\begin{align} & {{x}^{2}}-3x \\ & \text{For perfect square }{{b}^{2}}=4ac \\ & \therefore {{x}^{2}}-3x+c \\ & a=1,\text{ }b=-3,\text{ }c=k \\ & {{(-3)}^{2}}=4(1)(k) \\ & 9=4k \\ & k=\tfrac{9}{4} \\\end{align}$
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