Find the possible set of values of x for which expansion of `(3-2x)^(1//2)` is valid in ascending powers of x.

Answer

To find the possible set of values of \( x \) for which the expansion of \( (3 - 2x)^{1/2} \) is valid in ascending powers of \( x \), we can follow these steps:

### Step 1: Identify the form of the binomial expansion
The binomial expansion for \( (1 + u)^n \) is valid when \( |u| < 1 \). Here, we need to express \( (3 - 2x)^{1/2} \) in a suitable form.

### Step 2: Rewrite the expression
We can factor out \( 3 \) from \( (3 - 2x)^{1/2} \):
\[ (3 - 2x)^{1/2} = \sqrt{3} \left(1 - \frac{2}{3}x\right)^{1/2} \]

### Step 3: Apply the binomial expansion condition
For the expansion to be valid, we require:
\[ \left| -\frac{2}{3}x \right| < 1 \]
This simplifies to:
\[ \frac{2}{3}|x| < 1 \]

### Step 4: Solve the inequality
To solve the inequality \( \frac{2}{3}|x| < 1 \), we first multiply both sides by \( \frac{3}{2} \):
\[ |x| < \frac{3}{2} \]

### Step 5: Write the final range for \( x \)
The inequality \( |x| < \frac{3}{2} \) implies:
\[ -\frac{3}{2} < x < \frac{3}{2} \]

### Final Answer
Thus, the possible set of values of \( x \) for which the expansion is valid is:
\[ x \in \left(-\frac{3}{2}, \frac{3}{2}\right) \]

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