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The problem asks to determine all real values of \( a \) such that the equation \( ax^2 + 13x + 1 = 0 \) admits two distinct solutions.

### Approach:
For a quadratic equation \( ax^2 + bx + c = 0 \) to have two distinct real solutions, its discriminant must be strictly positive. The discriminant \( \Delta \) for a quadratic equation is given by:

\[
\Delta = b^2 - 4ac
\]

For the given equation \( ax^2 + 13x + 1 = 0 \), the coefficients are:
- \( a = a \) (unknown parameter),
- \( b = 13 \),
- \( c = 1 \).

Now, calculate the discriminant:

\[
\Delta = 13^2 - 4 \cdot a \cdot 1 = 169 - 4a
\]

For two distinct real solutions, the discriminant must be positive:

\[
169 - 4a > 0
\]

### Solving for \( a \):
\[
169 > 4a
\]
\[
a < \frac{169}{4}
\]
\[
a < 42.25
\]

Thus, the quadratic equation \( ax^2 + 13x + 1 = 0 \) will have two distinct real solutions for all values of \( a \) such that:

\[
a < 42.25
\]

Would you like more details or have any questions? Here are some follow-up questions for further exploration:

1. How do we solve the equation if \( a = 42.25 \)?
2. What happens if \( a = 0 \)?
3. How can we derive the general solution for a quadratic equation?
4. Can the discriminant ever be negative for this equation?
5. How does the value of \( a \) affect the shape of the parabola represented by the equation?

**Tip:** The discriminant is a key concept for determining the nature of solutions in quadratic equations. It can tell you whether the solutions are real or complex, and whether they are distinct or identical.

Déterminer tous les réels a tels que l'équation ax^2 + 13x + 1 = 0 admette deux solutions distinctes.

Learn how to determine all real values of a such that the quadratic equation ax^2 + 13x + 1 = 0 admits two distinct solutions. This problem covers key algebraic concepts including the discriminant and its application in quadratic equations. Suitable for students in Grades 10-12.

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