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 :

(A) $\mathrm{x}^{2}-35 \mathrm{x}+306=0$

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Statistics

Let the mean and the variance of seven observations

2, 4, α, 8, β, 12, 14, α < β

, be 8 and 16 respectively. Then the quadratic equation whose roots are

3 α + 2

and

2 β + 1

is :

Answer: (B)

The mean gives

2 + 4 + α + 8 + β + 12 + 14 = 56

, so

α + β = 16

. The variance gives

\frac{\sum x _{i}^{2}}{7} - 64 = 16

, so

\sum x_{i}^{2} = 560

, and since

2^{2} + 4^{2} + 8^{2} + 1 2^{2} + 1 4^{2} = 424

α^{2} + β^{2} = 136

. Then

2 α β = (α + β)^{2} - (α^{2} + β^{2}) = 256 - 136 = 120

, so

α β = 60

; with

α < β

this gives

α = 6, β = 10

. The required roots are

3 α + 2 = 20

and

2 β + 1 = 21

, so the quadratic is

x^{2} - (20 + 21) x + 20 \cdot 21 = x^{2} - 41 x + 420 = 0

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option