3. Quadratic Equations

A quadratic equation has the form \(ax^2 + bx + c = 0\) with \(a \neq 0\).

Method 1 - Factoring: Rewrite as a product that equals zero, then set each factor to zero.

Solve \(x^2 - 5x + 6 = 0\). Find two numbers that multiply to \(+6\) and add to \(-5\): those are \(-2\) and \(-3\).
\[ x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \]
So \(x - 2 = 0\) or \(x - 3 = 0\), giving \(x = 2\) or \(x = 3\).

Method 2 - The Quadratic Formula (always works):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For \(x^2 + 5x + 6 = 0\): \(a = 1,\, b = 5,\, c = 6\).
\[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)} = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm 1}{2} \]
So \(x = \frac{-4}{2} = -2\) or \(x = \frac{-6}{2} = -3\).

The discriminant \(b^2 - 4ac\) tells you how many real solutions exist: positive -> two, zero -> one, negative -> none.