JEE Main 2026 — Mathematics Probability & Statistics

Given the mean of the seven observations is $8$8:

$$\dfrac{2 + 4 + \alpha + 8 + \beta + 12 + 14}{7} = 8$$72+4+α+8+β+12+14​=8

$$\dfrac{40 + \alpha + \beta}{7} = 8$$740+α+β​=8

$$\alpha + \beta = 16$$α+β=16

Given the variance of the observations is $16$16:

$$\dfrac{\sum x_i^2}{n} - (\text{Mean})^2 = 16$$n∑xi2​​−(Mean)2=16

$$\dfrac{2^2 + 4^2 + \alpha^2 + 8^2 + \beta^2 + 12^2 + 14^2}{7} - 8^2 = 16$$722+42+α2+82+β2+122+142​−82=16

$$\dfrac{4 + 16 + \alpha^2 + 64 + \beta^2 + 144 + 196}{7} - 64 = 16$$74+16+α2+64+β2+144+196​−64=16

$$\dfrac{424 + \alpha^2 + \beta^2}{7} = 80$$7424+α2+β2​=80

$$424 + \alpha^2 + \beta^2 = 560$$424+α2+β2=560

$$\alpha^2 + \beta^2 = 136$$α2+β2=136

Using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$(α+β)2=α2+β2+2αβ:

$$16^2 = 136 + 2\alpha\beta$$162=136+2αβ

$$256 = 136 + 2\alpha\beta$$256=136+2αβ

$$2\alpha\beta = 120 \Rightarrow \alpha\beta = 60$$2αβ=120⇒αβ=60

Solving for $\alpha$α and $\beta$β, they are the roots of the equation $t^2 - 16t + 60 = 0$t2−16t+60=0, which are $6$6 and $10$10.

Since $\alpha < \beta$α<β, we have $\alpha = 6$α=6 and $\beta = 10$β=10.

The roots of the required quadratic equation are $3\alpha + 2$3α+2 and $2\beta + 1$2β+1:

$$3\alpha + 2 = 3(6) + 2 = 20$$3α+2=3(6)+2=20

$$2\beta + 1 = 2(10) + 1 = 21$$2β+1=2(10)+1=21

Sum of the roots $= 20 + 21 = 41$=20+21=41

Product of the roots $= 20 \times 21 = 420$=20×21=420

The required quadratic equation is $x^2 - (\text{Sum})x + (\text{Product}) = 0$x2−(Sum)x+(Product)=0:

$$x^2 - 41x + 420 = 0$$x2−41x+420=0

Answer: $x^2 - 41x + 420 = 0$x2−41x+420=0