Completing the Square and Vertex Form
Completing the square turns x2 + bx + c into a perfect square binomial by adding (b/2)2 to both sides, which solves the equation and reveals the vertex form a(x-h)2 + k at the same time.
A quadratic like x2 + 6x is not yet a perfect square, but it is one missing piece away from it. Picture x2 + 6x as an incomplete square of area: a square of side x plus two rectangles of width 3. To close the corner you need a 3-by-3 square, area 9. That 9 is exactly (half of 6) squared. Add it to both sides and the left side becomes (x+3)2, a true perfect square you can take the square root of directly.
The rule generalizes: for x2 + bx, add (b/2)2 to both sides. For x2 + 6x - 7 = 0, move the -7 over to get x2 + 6x = 7, add 9 to both sides to get x2 + 6x + 9 = 16, which factors as (x+3)2 = 16. Take square roots: x + 3 = ±4, so x = 1 or x = -7.
The quadratic formula always works, so completing the square is not about solving power, it is about what the process leaves behind. Every step builds toward vertex form, a(x-h)2 + k, where (h,k) is the parabola's turning point. The quadratic formula hands you roots and nothing else; completing the square hands you the vertex, the axis of symmetry, and the roots all from one rewritten equation. That makes it the bridge between solving equations and graphing parabolas, which is exactly why it keeps showing up alongside both skills.
Key terms
- Vertex form
- A quadratic written as a(x-h)2 + k, where (h,k) is the parabola's highest or lowest point.
- Perfect square binomial
- An expression like (x+3)2 that expands to x2 plus twice the product plus a squared constant.
- Axis of symmetry
- The vertical line x = h through the vertex that splits the parabola into mirror-image halves.
- Leading coefficient
- The number multiplying x2, which must be factored out first before completing the square if it isn't 1.
Watch it explained
- How it worksHow does half the x-coefficient, then squared, fill in the missing corner of a square of area?
- Worked exampleWalk me through solving x2 + 6x - 7 = 0 by completing the square.
- Common mistakeWhy isn't completing the square wasted effort when the quadratic formula always works?
- Going furtherWhat changes about completing the square when the leading coefficient is not 1?
Check yourself
End of topic.