Given that $x = -4$ and $x = -3$ are the roots of the equation $ax^2 + 7x + b = 0$, we can use Vieta's formulas, which state that for a quadratic equation $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$:

1. The sum of the roots $r_1 + r_2 = -\frac{b}{a}$.
2. The product of the roots $r_1 \cdot r_2 = \frac{c}{a}$.

Step 1: Sum of the roots

For this equation, the sum of the roots is: $-4 + (-3) = -7$ Using Vieta's formula for the sum of the roots: $-\frac{7}{a} = -7$ Solving for $a$: $a = 1$

Step 2: Product of the roots

The product of the roots is: $-4 \cdot -3 = 12$ Using Vieta's formula for the product of the roots: $\frac{b}{a} = 12$ Since $a = 1$, we have: $b = 12$

Conclusion

The values of $a$ and $b$ are: $a = 1 \quad \text{and} \quad b = 12$

Would you like further details on the steps? Here are some related questions:

1. How would the values of $a$ and $b$ change if the roots were different?
2. What happens if $a$ is not equal to 1?
3. How does Vieta's formula apply to equations of higher degrees?
4. Can Vieta's formulas be used if the roots are complex numbers?
5. What is the significance of the coefficient values in a quadratic equation?

Tip: Always verify your answers by plugging the roots back into the equation to check if it equals zero.