Solve `sqrt((x)/(x-3))+sqrt((x-3)/(x))=(5)/(2)` (`x != 0, x != 3`)

Answer

To solve the equation

\[ \sqrt{\frac{x}{x-3}} + \sqrt{\frac{x-3}{x}} = \frac{5}{2} \]

where \(x \neq 0\) and \(x \neq 3\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:

\[ \sqrt{\frac{x}{x-3}} + \sqrt{\frac{x-3}{x}} = \frac{5}{2} \]

### Step 2: Take LCM
To combine the square roots on the left side, we take the LCM. The LCM of the denominators \(x\) and \(x-3\) is \(\sqrt{x} \cdot \sqrt{x-3}\). Thus, we rewrite the left side:

\[ \frac{\sqrt{x} \cdot \sqrt{x} + \sqrt{x-3} \cdot \sqrt{x-3}}{\sqrt{x} \cdot \sqrt{x-3}} = \frac{5}{2} \]

This simplifies to:

\[ \frac{\sqrt{x}^2 + \sqrt{x-3}^2}{\sqrt{x} \cdot \sqrt{x-3}} = \frac{5}{2} \]

### Step 3: Simplify the numerator
The numerator simplifies to:

\[ \frac{x + (x - 3)}{\sqrt{x(x-3)}} = \frac{2x - 3}{\sqrt{x(x-3)}} \]

So now we have:

\[ \frac{2x - 3}{\sqrt{x(x-3)}} = \frac{5}{2} \]

### Step 4: Cross-multiply
Cross-multiplying gives us:

\[ 2(2x - 3) = 5\sqrt{x(x-3)} \]

This simplifies to:

\[ 4x - 6 = 5\sqrt{x(x-3)} \]

### Step 5: Square both sides
Next, we square both sides to eliminate the square root:

\[ (4x - 6)^2 = (5\sqrt{x(x-3)})^2 \]

Expanding both sides results in:

\[ 16x^2 - 48x + 36 = 25x(x - 3) \]

### Step 6: Expand the right side
Expanding the right side gives:

\[ 16x^2 - 48x + 36 = 25x^2 - 75x \]

### Step 7: Rearrange the equation
Rearranging all terms to one side results in:

\[ 16x^2 - 25x^2 + 75x - 48x + 36 = 0 \]

This simplifies to:

\[ -9x^2 + 27x + 36 = 0 \]

### Step 8: Multiply through by -1
To make the leading coefficient positive, we multiply through by -1:

\[ 9x^2 - 27x - 36 = 0 \]

### Step 9: Factor the quadratic
We can factor this quadratic equation:

\[ 9(x^2 - 3x - 4) = 0 \]

Factoring further gives:

\[ 9(x - 4)(x + 1) = 0 \]

### Step 10: Solve for \(x\)
Setting each factor to zero gives us:

\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Final Solution
Thus, the solutions to the equation are:

\[ x = 4 \quad \text{and} \quad x = -1 \]