Quadratic equations
Everyone can recite “minus b plus or minus the square root of delta, all over two a”, and hardly anyone knows where that line comes from. It is nothing more than an algebraic identity: the one you get by completing the square. Once you have seen it, it explains the discriminant, the sum and product of the roots, and the vertex of the parabola.
0What we are looking for
A quadratic equation is written ax² + bx + c = 0, where a is not 0. Solving it means finding where the parabola y = ax² + bx + c crosses the x-axis.
| Piece | Role |
|---|---|
| a | The coefficient of x², never zero. Positive, it opens the parabola upwards; negative, downwards. |
| b | The coefficient of x, which shifts the vertex to the left or to the right. |
| c | The constant term: the parabola crosses the vertical axis at (0, c). |
| Δ | The discriminant b² − 4ac, which decides whether there are two roots, one or none. |
1Completing the square
The formula does not come out of nowhere. You rewrite ax² + bx + c as a perfect square plus a constant, which lets you isolate x.
In the form ax² + bx + c, x appears twice and cannot be isolated. In the form a(x − α)² + β, it appears only once, inside a square: all that is left is to take a square root. The whole task is getting from one form to the other.
The equation becomes a(x + b/2a)² = Δ/4a, with Δ = b² − 4ac. Divide by a, take the square root of both sides, then isolate x. The ± sign appears because a square has two square roots, one positive and one negative.
2What Δ decides
In a(x + b/2a)² = Δ/4a, the left-hand side is a square multiplied by a. The sign of Δ tells you whether the equality can hold at all.
| Sign of Δ | Real roots | Example |
|---|---|---|
| Δ > 0 | two, distinct | x² − 5x + 6 = 0: Δ = 1, roots 2 and 3 |
| Δ = 0 | one, repeated | x² − 6x + 9 = 0: Δ = 0, root 3 |
| Δ < 0 | none | x² + 2x + 5 = 0: Δ = −16 |
No real number has a negative square: if Δ is negative, the equality is impossible. If Δ equals 0, the ± sign no longer makes any difference and the single root −b/2a is called a repeated root. The vertex of the parabola then sits right on the axis.
3Sum and product of the roots
When Δ is positive, the roots x₁ and x₂ satisfy two relations that often let you avoid the square root altogether.
Adding the two roots given by the formula makes the ± vanish: (−b + √Δ)/2a + (−b − √Δ)/2a = −b/a. Multiplying them gives (b² − Δ)/4a² = 4ac/4a² = c/a. For 2x² + 3x − 2 = 0, the roots 1/2 and −2 do indeed give −3/2 and −1.
4Vertex form and vertex
Completing the square also delivers the vertex form, and with it the coordinates of the vertex.
The square (x − α)² vanishes at x = α: the vertex is the point (α, β), a minimum if a is positive, a maximum otherwise. For x² − 4x + 1, α = 2 and β = −3, that is (x − 2)² − 3.
You can reach α another way: the derivative 2ax + b vanishes exactly at x = −b/2a, where the tangent is horizontal (see derivatives and antiderivatives).
5The lab
Vary a, b and c: the discriminant, the roots and the vertex are recomputed, and the parabola is redrawn.
6The traps
7Test yourself
Five questions, one correct answer each time.
8Glossary
| Term | Meaning |
|---|---|
| discriminant | the number Δ = b² − 4ac, whose sign fixes the number of real roots |
| root | a value of x that makes the expression zero, in other words a solution of the equation |
| repeated root | the single root when Δ = 0, equal to −b/2a; also called a double root |
| completing the square | rewriting ax² + bx + c as a square plus a constant |
| vertex form | the form a(x − α)² + β, which gives the vertex; called the canonical form in French textbooks |
| vertex | the point (α, β), the minimum or maximum of the parabola |
| Vieta’s formulas | x₁ + x₂ = −b/a and x₁ · x₂ = c/a |
| quadratic formula | x = (−b ± √Δ) / 2a; German pupils call it the Mitternachtsformel, the midnight formula |