JEE Main 2026 — 23 January, Morning Shift — Question 8

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

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

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

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

$\Rightarrow \mathrm{x}=1 \text { or } 5$

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

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

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

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

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

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

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

$\Rightarrow \mathrm{x}=25$

Now let $0<\mathrm{x}<9$, $-(\sqrt{3}-2)(t-3)+(t-3)^{2}-2 \sqrt{3}=0$

let $\mathrm{t}-3=\mathrm{u}$

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

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

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

$\mathrm{t}-3=-2$

$\Rightarrow \mathrm{t}=1 \Rightarrow \mathrm{x}=1$

$\alpha=1, \beta=25$

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