If `x^2 + 1/5 x+ a^2 `is a perfect square, then a is.
यदि `x^2 + 1/5 x+ a^2 ` एक पूर्ण वर्ग हे a तो का मान हे

Options

A. `1/100`
B. `pm1/10`
C. `1/10`
D. 10

Explanation

To determine the value of \( a \) such that the expression \( x^2 + \frac{1}{5}x + a^2 \) is a perfect square, we can follow these steps:

### Step 1: Identify the structure of a perfect square trinomial
A perfect square trinomial can be expressed in the form:
\[ (a + b)^2 = a^2 + 2ab + b^2 \]
In our case, we need to match the expression \( x^2 + \frac{1}{5}x + a^2 \) to this form.

### Step 2: Compare coefficients
From the expression \( x^2 + \frac{1}{5}x + a^2 \), we can identify:
- The coefficient of \( x^2 \) is \( 1 \) (which corresponds to \( a^2 \)).
- The coefficient of \( x \) is \( \frac{1}{5} \) (which corresponds to \( 2ab \)).
- The constant term is \( a^2 \) (which corresponds to \( b^2 \)).

### Step 3: Set up equations
From the coefficient of \( x \):
\[ 2ab = \frac{1}{5} \]
Let \( b = a \) (since we are looking for the same \( a \) in both terms):
\[ 2a^2 = \frac{1}{5} \]

### Step 4: Solve for \( a^2 \)
Now, we can solve for \( a^2 \):
\[ a^2 = \frac{1}{10} \]

### Step 5: Find \( a \)
Taking the square root of both sides gives:
\[ a = \pm \sqrt{\frac{1}{10}} = \pm \frac{1}{\sqrt{10}} = \pm \frac{1}{10} \]

### Conclusion
Thus, the value of \( a \) is:
\[ a = \frac{1}{10} \text{ or } -\frac{1}{10} \]