https://file.aiquickdraw.com/mathbot/file/1727325578379dappl3q1.jpg

The inequality shown in the image is:

\[
1 + q^2 > 0
\]

### Solution:

- The term \( q^2 \) represents the square of the variable \( q \). Since squaring any real number results in a non-negative value (i.e., \( q^2 \geq 0 \) for all \( q \in \mathbb{R} \)), this means that \( q^2 \) is always either zero or positive.
  
- Adding 1 to \( q^2 \) guarantees that the expression \( 1 + q^2 \) is strictly greater than zero, no matter what value \( q \) takes. Even if \( q = 0 \), the inequality holds because:

  \[
  1 + 0^2 = 1 > 0
  \]

### Conclusion:
The inequality \( 1 + q^2 > 0 \) is always true for all real values of \( q \). Therefore, the solution is:

\[
q \in \mathbb{R} \quad \text{(all real numbers)}
\]

Would you like further clarification or have any questions?

Here are 5 related questions you might want to explore:
1. How would the inequality change if it were \( 1 - q^2 > 0 \)?
2. What would the solution be if the inequality were \( 1 + q^2 \geq 0 \) instead?
3. Can you think of a situation where \( q^2 \) could be negative?
4. How would this inequality change if the constant 1 were replaced with another constant (e.g., \( 1 + q^2 > 5 \))?
5. What happens when \( q \) is a complex number?

**Tip:** Squaring any real number results in a non-negative value, which is useful to remember when solving inequalities involving squares.

Learn why the inequality 1 + q^2 > 0 is always true for all real values of q. This solution explains key concepts in quadratic functions and inequalities, ideal for students in Grades 9-12.

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