Quadratic Equations – Roots and Coefficients – JEE Main 23 Jan 2026 Shift 1

Solution:

Let the given equation be:
$$ (\sqrt{3}-2)|\sqrt{x}-3| + (x-6\sqrt{x}+9) – 2\sqrt{3} = 0 $$

Notice that $x-6\sqrt{x}+9 = (\sqrt{x}-3)^2 = |\sqrt{x}-3|^2$.
Let $u = |\sqrt{x}-3|$. The equation becomes:
$$ (\sqrt{3}-2)u + u^2 – 2\sqrt{3} = 0 $$
$$ u^2 – (2-\sqrt{3})u – 2\sqrt{3} = 0 $$

We factorize the quadratic in $u$:
$$ u^2 – 2u + \sqrt{3}u – 2\sqrt{3} = 0 $$
$$ u(u-2) + \sqrt{3}(u-2) = 0 $$
$$ (u-2)(u+\sqrt{3}) = 0 $$

So, $u = 2$ or $u = -\sqrt{3}$.
Since $u = |\sqrt{x}-3|$ is an absolute value, it must be non-negative.
Therefore, $u = 2$ is the only valid solution.

$$ |\sqrt{x}-3| = 2 $$
$$ \sqrt{x}-3 = 2 \quad \text{or} \quad \sqrt{x}-3 = -2 $$
$$ \sqrt{x} = 5 \quad \text{or} \quad \sqrt{x} = 1 $$
$$ x = 25 \quad \text{or} \quad x = 1 $$

Given $\alpha < \beta$, we have $\alpha = 1$ and $\beta = 25$.
We need to find $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$:
$$ \sqrt{\frac{25}{1}} + \sqrt{1 \cdot 25} $$
$$ = \sqrt{25} + \sqrt{25} $$
$$ = 5 + 5 = 10 $$

Ans. (3)