Aspire Faculty ID #17993 · Topic: JEE Main 2026 (23 January Morning Shift) · Just now
JEE Main 2026 (23 January Morning Shift)

If $\alpha$ and $\beta$ $(\alpha < \beta)$ are the roots of the equation


$(-2+\sqrt{3})\left(|\sqrt{x}-3|\right) + (x-6\sqrt{x}) + (9-2\sqrt{3}) = 0,\quad x \ge 0,$


then $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$ is equal to:

Solution

$(x-6\sqrt{x}+9) - (2-\sqrt{3})|\sqrt{x}-3| - 2\sqrt{3} = 0$

$\Rightarrow |\sqrt{x}-3|^2 - (2-\sqrt{3})|\sqrt{x}-3| - 2\sqrt{3} = 0$

$\Rightarrow |\sqrt{x}-3| = 2$ or $|\sqrt{x}-3| = -\sqrt{3}$ (not possible)

$\Rightarrow \sqrt{x} = 1$ or $5$

$\Rightarrow x = 1$ or $25$

$\Rightarrow \alpha = 1$ and $\beta = 25$

Aliter:

Let $x \ge 9$, let $\sqrt{x} = t \Rightarrow t \ge 3$

$(\sqrt{3}-2)(t-3) + (t-3)^2 - 2\sqrt{3} = 0$

let $t-3 = u$

$u^2 + (\sqrt{3}-2)u - 2\sqrt{3} = 0$

$u = 2$, or $u = -\sqrt{3}$

$\Rightarrow t-3 = 2$ or $t-3 = -\sqrt{3}$

$\Rightarrow t=5$ or $t=3-\sqrt{3}$ (rejected)

$\Rightarrow x=25$

Now let $0 < x < 9$,

$-(\sqrt{3}-2)(t-3) + (t-3)^2 - 2\sqrt{3} = 0$

let $t-3 = u$

$u^2 - (\sqrt{3}-2)u - 2\sqrt{3} = 0$

$\Rightarrow u = \sqrt{3}$ or $u=-2$

$\Rightarrow t=3+\sqrt{3}$ (rejected) or $t=1$

$\Rightarrow x=1$

$\alpha=1,; \beta=25$

Now $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta} = \sqrt{25} + \sqrt{25} = 10$
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