For what real values of a, will the expression `x^(2)-ax+1-2a^(2)`, for the real x, be always positive ?

Answer

To determine the values of \( a \) for which the expression \( x^2 - ax + 1 - 2a^2 \) is always positive for real \( x \), we need to analyze the quadratic expression in terms of its discriminant.

### Step-by-step Solution:

1. **Identify the Quadratic Expression**:
The expression is given as:
\[ f(x) = x^2 - ax + (1 - 2a^2) \]

2. **Condition for Always Positive**:
For the quadratic \( f(x) \) to be always positive, its discriminant must be less than zero. The discriminant \( D \) for a quadratic \( Ax^2 + Bx + C \) is given by:
\[ D = B^2 - 4AC \]
Here, \( A = 1 \), \( B = -a \), and \( C = 1 - 2a^2 \).

3. **Calculate the Discriminant**:
Substitute \( A \), \( B \), and \( C \) into the discriminant formula:
\[ D = (-a)^2 - 4 \cdot 1 \cdot (1 - 2a^2) \]
Simplifying this gives:
\[ D = a^2 - 4(1 - 2a^2) = a^2 - 4 + 8a^2 = 9a^2 - 4 \]

4. **Set the Discriminant Less Than Zero**:
We require:
\[ 9a^2 - 4 < 0 \]

5. **Solve the Inequality**:
Rearranging gives:
\[ 9a^2 < 4 \]
Dividing both sides by 9:
\[ a^2 < \frac{4}{9} \]

6. **Take the Square Root**:
Taking the square root of both sides, we get:
\[ -\frac{2}{3} < a < \frac{2}{3} \]

7. **Final Answer**:
Therefore, the values of \( a \) for which the expression is always positive are:
\[ a \in \left(-\frac{2}{3}, \frac{2}{3}\right) \]