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

$\frac{1}{3}$

$k(21x^2 + 24) + rx + (14x^2 − 9) = 0$ ----- (eq.1)

$k(7x^ 2 + 8) + px + (2x^ 2 − 3) = 0$ ------ (eq. 2)

Since the roots of the equation are common, the coefficients are in proportion.

The eq.1 can be written as, (21k +14)x² + rx + 24k - 9 = 0

The eq.2 can be written as, (7k + 2)x² + px + 8k - 3 = 0

= \(\frac{21k + 14}{7k + 2}\) = \(\frac{r}{p}\) = \(\frac{24k - 9}{8k - 3}\) 

= \(\frac{r}{p}\) = \(\frac{24k - 9}{8k - 3}\) 

= \(\frac{r}{p}\) = 3 × \(\frac{8k - 3}{8k - 3}\) 

= \(\frac{r}{p}\) = \(\frac{3}{1}\) 

= \(\frac{p}{r}\) = \(\frac{1}{3}\)