Question 16

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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: 

\[\begin{align}  & (x-\tfrac{1}{2})(x+\tfrac{2}{3})=0 \\ & {{x}^{2}}+\tfrac{2}{3}x-\tfrac{1}{2}x-\tfrac{1}{3}=0 \\ & \text{Multiply through by }6 \\ & 6{{x}^{2}}+4x-3x-2=0 \\ & 6{{x}^{2}}+x-2=0 \\\end{align}\]

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