Let's proceed with part (a) first.

Part (a):

Given the quadratic equation: $x^2 + bx + c = 0$ where $x_1$ and $x_2$ are the solutions (roots) of this equation.

The general formula for the roots of a quadratic equation $ax^2 + bx + c = 0$ is: $x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $x^2 + bx + c = 0$, we have $a = 1$. Using the quadratic formula, the roots are: $x_1 = \frac{-b + \sqrt{b^2 - 4c}}{2} \quad \text{and} \quad x_2 = \frac{-b - \sqrt{b^2 - 4c}}{2}$

Sum of the Roots:

The sum $x_1 + x_2$ can be calculated as: $x_1 + x_2 = \frac{-b + \sqrt{b^2 - 4c}}{2} + \frac{-b - \sqrt{b^2 - 4c}}{2}$ $x_1 + x_2 = \frac{-b - b}{2} = \frac{-2b}{2} = -b$

Thus, $x_1 + x_2 = -b$.

Product of the Roots:

The product $x_1 \cdot x_2$ can be calculated as: $x_1 \cdot x_2 = \left(\frac{-b + \sqrt{b^2 - 4c}}{2}\right) \cdot \left(\frac{-b - \sqrt{b^2 - 4c}}{2}\right)$ This is a difference of squares: $x_1 \cdot x_2 = \frac{(-b)^2 - (\sqrt{b^2 - 4c})^2}{4} = \frac{b^2 - (b^2 - 4c)}{4} = \frac{4c}{4} = c$

Thus, $x_1 \cdot x_2 = c$.

Conclusion for Part (a):

We have shown that: $x_1 + x_2 = -b \quad \text{and} \quad x_1 \cdot x_2 = c$

Part (b):

Now, let's use the results from part (a) to solve the equation: $x^2 - 5x + 6 = 0$ Here, $b = -5$ and $c = 6$. According to the results: $x_1 + x_2 = -(-5) = 5 \quad \text{and} \quad x_1 \cdot x_2 = 6$

We need to find two integers $x_1$ and $x_2$ such that:

1. Their sum is 5.
2. Their product is 6.

The two integers that satisfy these conditions are $2$ and $3$. Thus: $x_1 = 2, \quad x_2 = 3$

Solution:

The solutions to the equation $x^2 - 5x + 6 = 0$ are: $x = 2 \quad \text{and} \quad x = 3$

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How do we determine the discriminant of a quadratic equation and what does it tell us?
2. Can the sum and product formulas be used to find roots without directly solving the quadratic formula?
3. How would the solutions change if $b = 0$ in the equation $x^2 + bx + c = 0$?
4. What is the relationship between the coefficients of a polynomial and the roots for polynomials of higher degrees?
5. How do we solve quadratic equations using factoring?

Tip: The sum and product of the roots approach can often be quicker than directly using the quadratic formula, especially if the roots are integers.