waecmaths question:
The roots of a quadratic equation are $-\frac{1}{2}$ and $\frac{2}{3}$ . Find the equation
Option A:
$6{{x}^{2}}-x+2=0$
Option B:
$6{{x}^{2}}-x-2=0$
Option C:
$6{{x}^{2}}+x-2=0$
Option D:
$6{{x}^{2}}+x+2=0$
waecmaths solution:
$\begin{align} & (x+\tfrac{1}{2})(x-\tfrac{2}{3})=0 \\ & {{x}^{2}}-\tfrac{2}{3}x+\tfrac{1}{2}x-\tfrac{1}{3}=0 \\ & \text{Multiply through by 6} \\ & 6{{x}^{2}}-4x+3x-2=0 \\ & 6{{x}^{2}}-x-2=0 \\\end{align}$
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