If the term free from x in the expansion of `(sqrt(x)-(k)/(x^(2)))^(10)` is 405, then find the value of k.

Answer

To find the value of \( k \) in the expression \( \left(\sqrt{x} - \frac{k}{x^2}\right)^{10} \) such that the term free from \( x \) is 405, we can follow these steps:

### Step 1: Identify the General Term
The general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, we have:
- \( n = 10 \)
- \( a = \sqrt{x} \)
- \( b = -\frac{k}{x^2} \)

Thus, the general term becomes:
\[ T_r = \binom{10}{r} (\sqrt{x})^{10-r} \left(-\frac{k}{x^2}\right)^r \]

### Step 2: Simplify the General Term
Now, we simplify \( T_r \):
\[ T_r = \binom{10}{r} (\sqrt{x})^{10-r} \left(-\frac{k^r}{x^{2r}}\right) \]
This can be rewritten as:
\[ T_r = \binom{10}{r} (-k)^r x^{\frac{10-r}{2} - 2r} \]
\[ = \binom{10}{r} (-k)^r x^{\frac{10 - r - 4r}{2}} = \binom{10}{r} (-k)^r x^{\frac{10 - 5r}{2}} \]

### Step 3: Find the Term Free from \( x \)
For the term to be free from \( x \), the exponent of \( x \) must be zero:
\[ \frac{10 - 5r}{2} = 0 \]
Multiplying both sides by 2 gives:
\[ 10 - 5r = 0 \]
Solving for \( r \):
\[ 5r = 10 \implies r = 2 \]

### Step 4: Substitute \( r \) into the General Term
Now we substitute \( r = 2 \) back into the expression for \( T_r \):
\[ T_2 = \binom{10}{2} (-k)^2 \]
Calculating \( \binom{10}{2} \):
\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]
Thus:
\[ T_2 = 45 k^2 \]

### Step 5: Set the Expression Equal to 405
According to the problem, this term is equal to 405:
\[ 45 k^2 = 405 \]

### Step 6: Solve for \( k^2 \)
Dividing both sides by 45:
\[ k^2 = \frac{405}{45} = 9 \]

### Step 7: Find \( k \)
Taking the square root of both sides:
\[ k = 3 \quad \text{(since \( k \) is positive)} \]

Thus, the value of \( k \) is:
\[ \boxed{3} \]