waecmaths question:
Solve $4{{x}^{2}}-16x+15=0$
Option A:
$x=1\tfrac{1}{2}$ or $x=-2\tfrac{1}{2}$
Option B:
$x=1\tfrac{1}{2}$or $x=2\tfrac{1}{2}$
Option C:
$x=1\tfrac{1}{2}$ or $x=2\tfrac{1}{2}$
Option D:
$x=-1\tfrac{1}{2}$ or $x=-2\tfrac{1}{2}$
waecmaths solution:
$\begin{align} & 4{{x}^{2}}-16x+15=0 \\ & 4{{x}^{2}}-10x-6x+15=0 \\ & 2x(2x-5)-3(2x-5)=0 \\ & (2x-3)(2x-5)=0 \\ & x=\frac{3}{2}\text{ or }x=\frac{5}{2} \\ & x=1\tfrac{1}{2}\text{ or }x=2\tfrac{1}{2} \\\end{align}$
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