Can you walk me through solving a quadratic equation?

A quadratic equation is any equation that can be rearranged into this standard form:

$ax^2 + bx + c = 0$

The goal is to find the values of $x$ that make the equation true. The most reliable way to solve any quadratic equation is the Quadratic Formula:

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

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Let's walk through an example: $x^2 - 5x + 6 = 0$

1. Identify your variables:

• $a = 1$ (the number in front of $x^2$)
• $b = -5$ (the number in front of $x$)
• $c = 6$ (the constant at the end)

2. Plug them into the formula:

$$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}$$

3. Simplify the parts:

• $-(-5)$ becomes $5$
• $(-5)^2$ is $25$
• $4(1)(6)$ is $24$
• $2(1)$ is $2$

So now we have:

$$x = \frac{5 \pm \sqrt{25 - 24}}{2}$$ $$x = \frac{5 \pm \sqrt{1}}{2}$$

4. Solve for the two possibilities: Since the square root of $1$ is $1$, we have:

• $x = \frac{5 + 1}{2} = \frac{6}{2} = \mathbf{3}$
• $x = \frac{5 - 1}{2} = \frac{4}{2} = \mathbf{2}$

The solutions are $x = 3$ and $x = 2$.

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A few things to keep in mind:

• The $\pm$ symbol: This is why you get two answers. One path uses addition, the other uses subtraction.
• The "Discriminant" ($b^2 - 4ac$):
  • If this part is positive, you get two real answers.
  • If it's zero, you get exactly one answer.
  • If it's negative, you'll be dealing with imaginary numbers (the square root of a negative), which is a different map entirely!

Does that make the path through the equation a bit clearer, matey?