Does completing the square always work for quadratic equations?
Let's discuss why solving quadratic equations by completing the square works!
Roots2 and 1
Sum of roots3always −b/a
Product of roots2always c/a
Discriminant b² − 4ac1positive — two real roots
Vertex(1.5, -0.25)
Axis of symmetryx = 1.5
Opensupwardsthe vertex is the minimum
Forms of the same equation
| Standard | x^2 - 3x + 2 = 0 |
| Completed square | (x − 1.5)² − 0.25 = 0 |
| Vertex form | y = (x − 1.5)² − 0.25 |
| Factored | 1(x − 2)(x − 1) |
The method
x = (−b ± √(b² − 4ac)) ÷ 2a
The discriminant decides everything
b² − 4ac is the part under the square root, so its sign determines the nature of the roots before you compute them:
- Positive — two distinct real roots; the parabola crosses the x-axis twice.
- Zero — one repeated root; the vertex sits on the axis.
- Negative — two complex conjugate roots; the curve never touches the axis.
Vieta’s shortcuts
The roots always sum to −b/a and multiply to c/a. That gives a fast sanity check on any answer, and it is how factoring by inspection works.