The number of solutions of the equation (9/x - 9/\sqrt{x} + 2)(2/x - 7/\sqrt{x} + 3) = 0 is:

Q: Why can't we just multiply by $x$ to clear the denominators?

You absolutely could, but it would lead to a quartic equation (a polynomial of degree 4), which is significantly harder to solve than two simple quadratic equations. The substitution method is the elegant, 'JEE-smart' way to handle this.

Q: Do we need to check for extraneous solutions?

Yes, always! In any equation involving square roots or denominators, you must ensure your final values of $x$ satisfy the domain constraints. Here, $x$ must be strictly greater than zero. Since all our solutions are positive, they are all valid.

Q: What if one of the $\alpha$ values was negative?

Great question! Since $\alpha = 1/\sqrt{x}$, and the principal square root of a positive number is always positive, $\alpha$ must be strictly positive. If we had found a negative value for $\alpha$, we would have had to reject it immediately as it would not yield a real solution for $x$.

Question

The number of solutions of the equation $(\frac{9}{x}-\frac{9}{\sqrt{x}}+2)(\frac{2}{x}-\frac{7}{\sqrt{x}}+3)=0$ is:

Options:

(A) 2

(B) 3

(C) 1

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical realm! Today, we stand before an equation that, at first glance, might seem like a tangled mess of fractions and square roots:

$$ \left( \frac{9}{x} - \frac{9}{\sqrt{x}} + 2 \right) \left( \frac{2}{x} - \frac{7}{\sqrt{x}} + 3 \right) = 0 $$

It looks intimidating, but in the world of JEE Advanced, intimidation is just a mask for a beautiful, hidden simplicity. Let's peel back that mask together.

### The Gatekeeper: Domain Constraints

Before we touch the algebra, we must respect the Gatekeeper: the domain. We see $x$ in the denominator and inside a square root.

This tells us immediately that $x$ cannot be zero, and it cannot be negative. Thus, our domain is $x > 0$. This is our safety net; any solution we find must pass through this gate.

### The Key: The Power of Substitution

Dealing with $1/\sqrt{x}$ and $1/x$ directly is like trying to solve a puzzle with missing pieces. Let's use a powerful tool: substitution.

Let us define a new variable, $\alpha$, such that $\alpha = 1/\sqrt{x}$. Squaring both sides, we get $\alpha^2 = 1/x$.

Suddenly, the equation transforms into a product of two elegant quadratics:

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

### The Two Paths: Solving the Quadratics

The **Zero Product Property** is our best friend here. It tells us that if the product of two terms is zero, then at least one of them must be zero.

**Path 1:** $9\alpha^2 - 9\alpha + 2 = 0$. By splitting the middle term, we find $(3\alpha - 1)(3\alpha - 2) = 0$, giving us $\alpha = 1/3$ and $\alpha = 2/3$.

**Path 2:** $2\alpha^2 - 7\alpha + 3 = 0$. Again, splitting the middle term, we get $(2\alpha - 1)(\alpha - 3) = 0$, leading to $\alpha = 1/2$ and $\alpha = 3$.

We have four potential values for $\alpha$: $1/3, 2/3, 1/2,$ and $3$. Since all are positive, they all pass our domain check.

### The Final Transformation: Returning to $x$

We are almost at the finish line! The question asks for the number of solutions for $x$, so we must reverse our substitution.

Given $1/\sqrt{x} = \alpha$, it follows that $\sqrt{x} = 1/\alpha$, or $x = 1/\alpha^2$. Let's calculate $x$ for each value:

*   For $\alpha = 1/3$, $x = 1/(1/3)^2 = 9$.
*   For $\alpha = 2/3$, $x = 1/(2/3)^2 = 9/4$.
*   For $\alpha = 1/2$, $x = 1/(1/2)^2 = 4$.
*   For $\alpha = 3$, $x = 1/3^2 = 1/9$.

All four values are distinct and positive. We have successfully navigated the labyrinth! The **total number of solutions is 4**.

Animated Solution for Mathematics - Quadratic Equations: The number of solutions of the equation $(\frac{9}{x} - \frac{9}{x} + 2) (\frac{2}{x} - \frac{7}{x} + 3) = 0$ is:

Select Answer:

Visualized Solution

Initial Observation & Domain

The Substitution Strategy

Transforming the Equation

Zero Product Property

Solving the First Quadratic

Solving the Second Quadratic

Consolidating Alpha Values

Back-substitution for $x$

Calculating Final Solutions

Final Count of Solutions

The Sigma Insight: Transformation of Equations

Analyzing the Setup

The Gatekeeper

The Key

The Two Paths

The Final Transformation

Similar Questions

The equation $e^{4 x} + 8 e^{3 x} + 13 e^{2 x} - 8 e^{x} + 1 = 0, x \in R$ has:

The number of real roots of the equation $e^{4 x} - e^{3 x} - 4 e^{2 x} - e^{x} + 1 = 0$ is equal to

Find number of real roots of equation $e^{4 x} + e^{3 x} - 4 e^{2 x} + e^{x} + 1 = 0$ is

The number of real roots of the equation, $e^{4 x} + e^{3 x} - 4 e^{2 x} + e^{x} + 1 = 0$ is:

If $α$ and $β (α < β)$ are the roots of the equation $(- 2 + 3) (∣ x - 3∣) + (x - 6 x) + (9 - 23) = 0, x \geq 0$ , then $\frac{β}{α} + α β$ is equal to :

The sum of all the real roots of the equation $∣ x - 2 ∣^{2} + ∣ x - 2∣ - 2 = 0$ is .........

The sum of the cubes of all the roots of the equation $x^{4} - 3 x^{3} - 2 x^{2} + 3 x + 1 = 0$ is ______.

Solve for $x : (5 + 26)^{x^{2} - 3} + (5 - 26)^{x^{2} - 3} = 10$

The product of the roots of the equation $9 x^{2} - 18∣ x ∣ + 5 = 0$ , is:

Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of the equation (x9​−x​9​+2)(x2​−x​7​+3)=0 is: