If the equation k(21x² + 24) + rx + (14x² - 9) = 0
k(7x² + 8) + px + (2x² - 3) = 0 have both roots common, then the value of $$\frac{p}{r}$$ is:

Solution(By Examveda Team)

$$\eqalign{ & k\left( {21{x^2} + 24} \right) + rx + \left( {14{x^2} - 9} \right) = 0\,......\left( {\text{i}} \right) \cr & k\left( {7{x^2} + 8} \right) + px + \left( {2{x^2} - 3} \right) = 0\,......\left( {{\text{ii}}} \right) \cr & {\text{from }}\left( {\text{i}} \right), \cr & 21k{x^2} + 24k + rx + 14{x^2} - 9 = 0 \cr & \left( {21k + 14} \right){x^2} + rx + \left( {24k - 9} \right) = 0\,......\left( {{\text{iii}}} \right) \cr & {\text{from }}\left( {{\text{ii}}} \right), \cr & 7k{x^2} + 8k + px + 2{x^2} - 3 = 0 \cr & \left( {7k + 2} \right){x^2} + px + \left( {8k - 3} \right) = 0\,......\left( {{\text{iv}}} \right) \cr & {\text{If roots are common then,}} \cr & \frac{{21k + 14}}{{7k + 2}} = \frac{r}{p} = \frac{{24k - 9}}{{8k - 3}} \cr & \Rightarrow \frac{r}{p} = \frac{{3\left( {8k - 3} \right)}}{{\left( {8k - 3} \right)}} \cr & \Rightarrow \frac{r}{p} = \frac{3}{1} \cr & \Rightarrow \frac{p}{r} = \frac{1}{3} \cr} $$