Let the mean and the variance of seven observations 2, 4, $$\alpha$$, 8, $$\beta$$, 12, 14, $$\alpha < \beta$$, be 8 and 16 respectively. Then the quadratic equation whose roots are $$3\alpha + 2$$ and $$2\beta + 1$$ is :

$$\mu = \frac{\sum x_i}{N} \quad \text{and} \quad \sigma^2 = \frac{\sum x_i^2}{N} - \mu^2$$

$$\frac{2 + 4 + \alpha + 8 + \beta + 12 + 14}{7} = 8$$

$$40 + \alpha + \beta = 56$$

$$\alpha + \beta = 16 \implies \beta = 16 - \alpha$$

$$\sigma^2 = \frac{\sum x_i^2}{7} - \mu^2 = 16$$

$$\frac{2^2 + 4^2 + \alpha^2 + 8^2 + \beta^2 + 12^2 + 14^2}{7} - 8^2 = 16$$

$$\frac{4 + 16 + \alpha^2 + 64 + \beta^2 + 144 + 196}{7} - 64 = 16$$

$$\frac{424 + \alpha^2 + \beta^2}{7} = 16 + 64 = 80$$

$$424 + \alpha^2 + \beta^2 = 560$$

$$\alpha^2 + \beta^2 = 136$$

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$: $$136 = (16)^2 - 2\alpha\beta$$

$$2\alpha\beta = 256 - 136 = 120 \implies \alpha\beta = 60$$

We now have $$\alpha + \beta = 16$$ and $$\alpha\beta = 60$$.

Since it is given that $$\alpha < \beta$$: $$\alpha = 6 \quad \text{and} \quad \beta = 10$$

The roots are $$r_1 = 3\alpha + 2 = 3(6) + 2 = 20$$ and $$r_2 = 2\beta + 1 = 2(10) + 1 = 21$$

Quadratic equation: $$x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$$

$$\implies x^2 - 41x + 420 = 0$$