JEE Main 2026 Apr 6 Shift 2, Mathematics Q8: Quadratic Equations

JEE Main2026Apr 6, Shift 2Mathematics

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

A
$x^{2} - 35 x + 306 = 0$
B
$x^{2} - 41 x + 420 = 0$
C
$x^{2} - 45 x + 506 = 0$
D
$x^{2} - 37 x + 342 = 0$

Solution

Mean. $\frac{2 + 4 + α + 8 + β + 12 + 14}{7} = 8$ $\Rightarrow$ $α + β = 16$ ...(i)

Variance. $\frac{\sum x _{i}^{2}}{7} - 64 = 16$ $\Rightarrow$ $\sum x_{i}^{2} = 560$ .

$2^{2} + 4^{2} + 8^{2} + 1 2^{2} + 1 4^{2} = 4 + 16 + 64 + 144 + 196 = 424$ , so $α^{2} + β^{2} = 136$ ...(ii)

From (i), $α β = \frac{( α + β ) ^{2} - ( α ^{2} + β ^{2} )}{2} = \frac{256 - 136}{2} = 60$ . With $\alpha < \beta$: $\alpha = 6,\,\beta = 10$.

Roots: $3 α + 2 = 20$ and $2 β + 1 = 21$ . Sum $= 41$ , product $= 420$ .

$∴$ $x^{2} - 41 x + 420 = 0$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q8: Quadratic Equations

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