Quadratic Equation Formed from Roots of Variance and Mean Data

Given seven observations: $2, 4, \alpha, 8, \beta, 12, 14$ with $\alpha < \beta$.

Step 1: Utilize the Mean Formula

The mean ($\bar{x}$) of the given observations is $8$. $\bar{x} = \frac{\sum x_i}{N}$ $\frac{2 + 4 + \alpha + 8 + \beta + 12 + 14}{7} = 8$ $40 + \alpha + \beta = 56$ $\alpha + \beta = 16 \quad \text{--- (1)}$

Step 2: Utilize the Variance Formula

The variance ($\sigma^2$) of the observations is $16$. $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$ $16 = \frac{2^2 + 4^2 + \alpha^2 + 8^2 + \beta^2 + 12^2 + 14^2}{7} - 8^2$ $16 = \frac{4 + 16 + \alpha^2 + 64 + \beta^2 + 144 + 196}{7} - 64$ $80 = \frac{424 + \alpha^2 + \beta^2}{7}$ $560 = 424 + \alpha^2 + \beta^2$ $\alpha^2 + \beta^2 = 136 \quad \text{--- (2)}$

Step 3: Solve for $\alpha$ and $\beta$

Using the algebraic identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$: $(16)^2 = 136 + 2\alpha\beta$ $256 = 136 + 2\alpha\beta$ $2\alpha\beta = 120 \implies \alpha\beta = 60$

Now, $\alpha$ and $\beta$ are the roots of the quadratic equation $t^2 - (\alpha + \beta)t + \alpha\beta = 0$: $t^2 - 16t + 60 = 0$ $(t - 6)(t - 10) = 0$ $t = 6 \quad \text{or} \quad t = 10$

Given $\alpha < \beta$, we have: $\alpha = 6 \quad \text{and} \quad \beta = 10$

Step 4: Form the Required Quadratic Equation

The roots of the target quadratic equation are $3\alpha + 2$ and $2\beta + 1$.

• First root ($r_1$): $r_1 = 3\alpha + 2 = 3(6) + 2 = 20$

• Second root ($r_2$): $r_2 = 2\beta + 1 = 2(10) + 1 = 21$

• Sum of the roots ($S$): $S = r_1 + r_2 = 20 + 21 = 41$

• Product of the roots ($P$): $P = r_1 \times r_2 = 20 \times 21 = 420$

Thus, the required quadratic equation is: $x^2 - Sx + P = 0$ $x^2 - 41x + 420 = 0$

Therefore, the correct option is B.