Using the Formula
To use the quadratic formula, you must identify \(a, b,\) and \(c\) from your equation. The equation must be set to zero.
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
The Discriminant (\(D = b^2 - 4ac\))
- If \(D > 0\): Two real solutions (roots).
- If \(D = 0\): One real solution (the vertex).
- If \(D < 0\): Two Complex solutions!
Worked Examples
Example 1: Basic Application
Solve: \(x^2 - 5x + 6 = 0\)
- \(a=1, b=-5, c=6\).
- \(x = \frac{5 \pm \sqrt{25 - 4(1)(6)}}{2} = \frac{5 \pm \sqrt{1}}{2}\).
- \(x = \frac{5+1}{2} = 3\) and \(x = \frac{5-1}{2} = 2\).
- Result: \(\{2, 3\}\)
Example 2: Messy Roots
Solve: \(2x^2 + 4x - 1 = 0\)
- \(a=2, b=4, c=-1\).
- \(x = \frac{-4 \pm \sqrt{16 - 4(2)(-1)}}{4} = \frac{-4 \pm \sqrt{24}}{4}\).
- \(\sqrt{24} = 2\sqrt{6}\).
- \(x = \frac{-4 \pm 2\sqrt{6}}{4} = -1 \pm \frac{\sqrt{6}}{2}\).
- Result: \(\{-1 - \frac{\sqrt{6}}{2}, -1 + \frac{\sqrt{6}}{2}\}\)
The Bridge to Quantum Mechanics
In Quantum Mechanics, the energy of a system is found by finding the "roots" of the Hamiltonian operator. Often, these equations are quadratic. When the discriminant is negative, we get complex roots. In Chapter 4 and 15, we will see that these complex roots correspond to States of Spin. If the discriminant of the universe's energy equations didn't allow for complex numbers, particles wouldn't have spin, and magnets wouldn't work. The discriminant tells you what kind of matter is possible.