Algebra
Quadratic equations and functions
The quadratic formula
20 practice questions
2 video lessons
Theory + worked examples
Theory
The quadratic formula solves any quadratic:
\[x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.\]
Identify \(a,b,c\) from \(ax^2+bx+c=0\) and substitute.
It always works, even when factoring is difficult.
The quadratic formula.
A worked example.
The formula:
\[x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Compute \(b^2-4ac\) first, then the rest.
How to use the quadratic formula
- Write the equation as \(ax^2+bx+c=0\).
- Identify \(a,b,c\).
- Substitute into the formula.
- Simplify, keeping \(\pm\).
Example 1 — Apply the formula
Solve \(x^2-5x+6=0\).
Solution
Use \(a=1,b=-5,c=6\).
| \(x\) | \(=\) | \(\dfrac{5\pm\sqrt{25-24}}{2}\) |
| \(=\) | \(\dfrac{5\pm1}{2}=3,\ 2\) |
Example 2 — Irrational roots
Solve \(x^2-2x-1=0\).
Solution
\(D=4+4=8\).
| \(x\) | \(=\) | \(\dfrac{2\pm\sqrt8}{2}\) |
| \(=\) | \(1\pm\sqrt2\) |
Example 3 — Leading coefficient
Solve \(2x^2+3x-2=0\).
Solution
\(a=2,b=3,c=-2\), \(D=9+16=25\).
| \(x\) | \(=\) | \(\dfrac{-3\pm5}{4}\) |
| \(=\) | \(\dfrac12\ \text{or}\ -2\) |
Example 4 — When to use it
When is the quadratic formula best?
Solution
When a quadratic doesn't factor easily — it always works.
Common pitfalls
Use the signs of \(a,b,c\) carefully — \((-5)^2=25\).
Divide everything by \(2a\), including \(-b\).
Keep \(\pm\) for both solutions.
Frequently asked questions
What is the quadratic formula?
\(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\).
When should you use it?
For any quadratic, especially when it doesn't factor.
What is under the root called?
The discriminant, \(b^2-4ac\).
Does it always work?
Yes — for every quadratic.
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