Given the quadratic equation $16x^2 + 12x + c = 0$, the roots of the equation are denoted by $\alpha$ and $\beta$. We are required to find all possible values of $c$.

Step 1: Use the sum and product of roots formulas

For a quadratic equation $ax^2 + bx + c = 0$, the sum and product of the roots can be given by: $\alpha + \beta = -\frac{b}{a}$ $\alpha \beta = \frac{c}{a}$ Here, $a = 16$, $b = 12$, and $c = c$ (unknown). Thus, we have: $\alpha + \beta = -\frac{12}{16} = -\frac{3}{4}$ $\alpha \beta = \frac{c}{16}$

Step 2: Discriminant condition for real roots

For the roots $\alpha$ and $\beta$ to be real, the discriminant $\Delta$ of the quadratic equation must be non-negative. The discriminant for the equation $16x^2 + 12x + c = 0$ is given by: $\Delta = b^2 - 4ac = 12^2 - 4 \times 16 \times c$ Simplifying, we get: $\Delta = 144 - 64c$ For the roots to be real, we require $\Delta \geq 0$, so: $144 - 64c \geq 0$ $144 \geq 64c$ $c \leq \frac{144}{64} = \frac{9}{4}$

Step 3: Conclusion

The value of $c$ must satisfy $c \leq \frac{9}{4}$ for the roots to be real.

Thus, all possible values of $c$ are $c \leq \frac{9}{4}$.

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Would you like more details on any part of this solution or have any further questions?

Here are some related questions you might find interesting:

1. How can we find the roots if $c$ is known?
2. What happens if $c > \frac{9}{4}$?
3. Can we derive a specific value of $c$ for which the roots are equal?
4. How does the discriminant determine the nature of the roots?
5. How would the solution change if $c$ were negative?

Tip: Always check the discriminant when dealing with quadratic equations to determine the nature of the roots.