The minimum value of 'a' for which `a^(2)x^(2)+(2a-1)x+1` is non-negative for any real x is __________

Answer

To find the minimum value of \( a \) for which the expression \( a^2 x^2 + (2a - 1)x + 1 \) is non-negative for any real \( x \), we need to analyze the quadratic expression.

### Step-by-Step Solution:

1. **Identify the Quadratic Form**:
The expression can be written in the standard form of a quadratic equation:
\[ f(x) = a^2 x^2 + (2a - 1)x + 1 \]
Here, \( A = a^2 \), \( B = 2a - 1 \), and \( C = 1 \).

2. **Condition for Non-Negativity**:
For the quadratic expression to be non-negative for all \( x \), the discriminant must be less than or equal to zero. The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C \) is given by:
\[ D = B^2 - 4AC \]
Therefore, we need:
\[ D \leq 0 \]

3. **Calculate the Discriminant**:
Substitute \( A \), \( B \), and \( C \) into the discriminant formula:
\[ D = (2a - 1)^2 - 4(a^2)(1) \]
Expanding this gives:
\[ D = (2a - 1)^2 - 4a^2 \]
\[ = 4a^2 - 4a + 1 - 4a^2 \]
\[ = -4a + 1 \]

4. **Set the Discriminant Less Than or Equal to Zero**:
Now, set the discriminant less than or equal to zero:
\[ -4a + 1 \leq 0 \]

5. **Solve for \( a \)**:
Rearranging the inequality gives:
\[ 1 \leq 4a \]
Dividing both sides by 4:
\[ \frac{1}{4} \leq a \]
or
\[ a \geq \frac{1}{4} \]

6. **Conclusion**:
The minimum value of \( a \) for which the expression is non-negative for any real \( x \) is:
\[ \boxed{\frac{1}{4}} \quad \text{or} \quad 0.25 \]