Algebra – Quadratic Equations – JEE Main 29 Jan 2025 Shift 1

Solution:

Let $\frac{1}{\sqrt{x}} = \alpha$. Since $x > 0$ for the terms to be defined, $\alpha > 0$.

Substituting $\alpha$ into the equation:

$(9\alpha^2 – 9\alpha + 2)(2\alpha^2 – 7\alpha + 3) = 0$

Now factorize each quadratic part:

Part 1: $9\alpha^2 – 9\alpha + 2 = 0$

$9\alpha^2 – 6\alpha – 3\alpha + 2 = 0$

$3\alpha(3\alpha – 2) – 1(3\alpha – 2) = 0$

$(3\alpha – 2)(3\alpha – 1) = 0$

$\alpha = \frac{2}{3}, \frac{1}{3}$. (Both are positive, so valid).

Part 2: $2\alpha^2 – 7\alpha + 3 = 0$

$2\alpha^2 – 6\alpha – \alpha + 3 = 0$

$2\alpha(\alpha – 3) – 1(\alpha – 3) = 0$

$(2\alpha – 1)(\alpha – 3) = 0$

$\alpha = \frac{1}{2}, 3$. (Both are positive, so valid).

The possible values for $\alpha = \frac{1}{\sqrt{x}}$ are $\{ \frac{1}{3}, \frac{2}{3}, \frac{1}{2}, 3 \}$.

Since each $\alpha$ corresponds to a unique $x = \frac{1}{\alpha^2}$, and all $\alpha$ values are distinct and positive, we get 4 distinct values of $x$.

$x \in \{ 9, \frac{9}{4}, 4, \frac{1}{9} \}$.

Total number of solutions is 4.

Ans. (2)