To complete Macey's work and solve the quadratic equation $x^2 - 3x - 2 = 0$, let's go through each step using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case:

• $a = 1$
• $b = -3$
• $c = -2$

Substitute these values into the formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}$

Simplify the terms:

1. Calculate $-(-3) = 3$.

   $x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}$

2. Evaluate $(-3)^2 = 9$ and $-4 \cdot 1 \cdot (-2) = 8$, so the expression inside the square root is:

   $9 + 8 = 17$

3. Substitute back:

   $x = \frac{3 \pm \sqrt{17}}{2}$

Thus, the solutions are:

$x = \frac{3 \pm \sqrt{17}}{2}$

Answer

The correct answer is: A. $\frac{3 \pm \sqrt{17}}{2}$

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Related Questions

1. How do you derive the quadratic formula?
2. How can you determine if a quadratic equation has real or complex roots?
3. What is the importance of the discriminant $b^2 - 4ac$ in quadratic equations?
4. How would the solutions change if $c$ was positive?
5. What are other methods to solve a quadratic equation?

Tip

Always check the discriminant ($b^2 - 4ac$) first; if it’s positive, the equation has two real solutions; if zero, one real solution; if negative, two complex solutions.