waecmaths question:
Form the equation whose roots are \[x=\tfrac{1}{2}\text{ and }-\tfrac{2}{3}\]
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:
\[\begin{align} & (x-\tfrac{1}{2})(x+\tfrac{2}{3})=0 \\ & {{x}^{2}}+\tfrac{2}{3}x-\tfrac{1}{2}x-\tfrac{1}{3}=0 \\ & \text{Multiply through by }6 \\ & 6{{x}^{2}}+4x-3x-2=0 \\ & 6{{x}^{2}}+x-2=0 \\\end{align}\]
maths year:
maths topics: