Let's start by solving the first few problems from question 2 using the quadratic formula and then verifying by factoring.

Question 2:

Solve each of the following using the quadratic formulas. Check by factoring.

(a) $x^2 - 7x + 12 = 0$

Quadratic Formula: The quadratic formula is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $x^2 - 7x + 12 = 0$, we have $a = 1$, $b = -7$, and $c = 12$.

1. Calculate the discriminant: $\Delta = b^2 - 4ac$ $\Delta = (-7)^2 - 4(1)(12)$ $\Delta = 49 - 48$ $\Delta = 1$

2. Apply the quadratic formula: $x = \frac{-(-7) \pm \sqrt{1}}{2(1)}$ $x = \frac{7 \pm 1}{2}$ $x_1 = \frac{7 + 1}{2} = 4$ $x_2 = \frac{7 - 1}{2} = 3$

Factoring: $x^2 - 7x + 12 = (x - 4)(x - 3) = 0$

Thus, the solutions are $x = 4$ and $x = 3$.

(b) $2x^2 + 5x + 3 = 0$

Quadratic Formula: For the equation $2x^2 + 5x + 3 = 0$, we have $a = 2$, $b = 5$, and $c = 3$.

1. Calculate the discriminant: $\Delta = b^2 - 4ac$ $\Delta = 5^2 - 4(2)(3)$ $\Delta = 25 - 24$ $\Delta = 1$

2. Apply the quadratic formula: $x = \frac{-5 \pm \sqrt{1}}{2(2)}$ $x = \frac{-5 \pm 1}{4}$ $x_1 = \frac{-5 + 1}{4} = \frac{-4}{4} = -1$ $x_2 = \frac{-5 - 1}{4} = \frac{-6}{4} = -\frac{3}{2}$

Factoring: $2x^2 + 5x + 3 = (2x + 3)(x + 1) = 0$

Thus, the solutions are $x = -1$ and $x = -\frac{3}{2}$.

(c) $x^2 - 4 = 0$

Quadratic Formula: For the equation $x^2 - 4 = 0$, we have $a = 1$, $b = 0$, and $c = -4$.

1. Calculate the discriminant: $\Delta = b^2 - 4ac$ $\Delta = 0^2 - 4(1)(-4)$ $\Delta = 16$

2. Apply the quadratic formula: $x = \frac{-0 \pm \sqrt{16}}{2(1)}$ $x = \frac{\pm 4}{2}$ $x_1 = \frac{4}{2} = 2$ $x_2 = \frac{-4}{2} = -2$

Factoring: $x^2 - 4 = (x - 2)(x + 2) = 0$

Thus, the solutions are $x = 2$ and $x = -2$.

(d) $x^2 - 5x = 0$

Quadratic Formula: For the equation $x^2 - 5x = 0$, we have $a = 1$, $b = -5$, and $c = 0$.

1. Calculate the discriminant: $\Delta = b^2 - 4ac$ $\Delta = (-5)^2 - 4(1)(0)$ $\Delta = 25$

2. Apply the quadratic formula: $x = \frac{-(-5) \pm \sqrt{25}}{2(1)}$ $x = \frac{5 \pm 5}{2}$ $x_1 = \frac{5 + 5}{2} = 5$ $x_2 = \frac{5 - 5}{2} = 0$

Factoring: $x^2 - 5x = x(x - 5) = 0$

Thus, the solutions are $x = 5$ and $x = 0$.

Would you like to continue with more problems, or do you have any questions about these solutions?

Here are 5 more questions you might want to ask:

1. Can you solve more problems from question 2?
2. How do you solve quadratic equations when the discriminant is negative?
3. What are some common mistakes to avoid when using the quadratic formula?
4. Can you explain the steps for factoring quadratic equations in more detail?
5. How do you determine the number of real roots of a quadratic equation?

Tip: Always check the discriminant ($b^2 - 4ac$) first when solving quadratic equations, as it tells you the nature of the roots (real and distinct, real and equal, or complex).