Question 10

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waecmaths question: 

The roots of a quadratic equation are$\tfrac{4}{3}$and $-\tfrac{3}{7}$. Find the equation. 

Option A: 

$21{{x}^{2}}-19x-12=0$

Option B: 

$21{{x}^{2}}+37x-12=0$

Option C: 

$21{{x}^{2}}-x+12=0$

Option D: 

$21{{x}^{2}}+7x-4=0$

waecmaths solution: 

$\begin{align}  & \left( x-\frac{4}{3} \right)\left( x+\frac{3}{7} \right)=0 \\ & {{x}^{2}}+\frac{3}{7}x+\frac{4}{3}x-\frac{12}{21}=0 \\ & \text{Multiply both sides by 21} \\ & 21x+9x-28x-12=0 \\ & 21{{x}^{2}}-19x-12=0 \\\end{align}$

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