The Completing the Square Formula

You need the general formula for completing the square and the exact derivation that turns it into the quadratic formula. Most algebra pages give you the vertex coordinates h = −b/2a and k = c − b²/4a, then stop. Here you get the full line-by-line derivation of the quadratic formula from completing the square, with the common failure points named and fixed.

The completing the square formula rewrites any quadratic f(x) = ax² + bx + c into vertex form f(x) = a(x − h)² + k.This form reveals the parabola's vertex, its axis of symmetry x = h, and it is the algebraic backbone of the quadratic formula, certain integrals, and circle equations.

Start with the general quadratic expression f(x) = ax² + bx + c, where a ≠ 0. The goal is to force a perfect square trinomial inside the parentheses. This derivation follows the standard algebra text notation from OpenStax College Algebra 2e, Section 5.1, pages 440-441.

Step 1: Factor a From the First Two Terms

Write f(x) = a[x² + (b/a)x] + c. Factoring a is critical when a ≠ 1. The most common failure here is to forget this step and try to complete the square directly on ax² + bx, which produces a wrong vertex.

Step 2: Add and Subtract (b/(2a))² Inside the Brackets

Inside the bracket, add and subtract the square of half the coefficient of x: (b/(2a))². The expression becomes a[x² + (b/a)x + (b/(2a))² − (b/(2a))²] + c. Adding and subtracting the same term keeps the expression balanced. This is the step where students who only ever solved equations mistakenly add (b/(2a))² to just one side. Because you are rewriting an expression, not solving an equation, you must add and subtract inside the same parentheses.

Step 3: Rewrite the Perfect Square Trinomial

The first three terms inside the bracket form a perfect square: x² + (b/a)x + (b/(2a))² = (x + b/(2a))². So now f(x) = a[(x + b/(2a))² − (b/(2a))²] + c.

Step 4: Distribute a and Simplify the Constant

Distribute a to get a(x + b/(2a))² − a(b/(2a))² + c. The term a(b/(2a))² simplifies to b²/(4a). So f(x) = a(x + b/(2a))² − b²/(4a) + c.

Step 5: Combine the Constants

Rewrite as f(x) = a(x + b/(2a))² + (c − b²/(4a)). Now replace + b/(2a) with −(−b/(2a)). Because the vertex form is a(x − h)², you get h = −b/(2a) and k = c − b²/(4a). The axis of symmetry is the vertical line x = h.

The quadratic formula x = [−b ± √(b² − 4ac)] / (2a) is derived directly from completing the square on the general equation ax² + bx + c = 0. This derivation is the single best proof of why the formula works, yet it is rarely shown to students. It takes eight lines. The source is OpenStax College Algebra 2e, Section 2.5, pages 218-219.

Begin with the equation ax² + bx + c = 0, a ≠ 0. Divide every term by a: x² + (b/a)x + c/a = 0. Isolate the x terms by subtracting c/a: x² + (b/a)x = −c/a. Now add (b/(2a))² to both sides. This is the key move: because you are solving an equation, you add the same quantity to both sides, not add and subtract inside one side. The equation becomes x² + (b/a)x + (b/(2a))² = −c/a + (b/(2a))².

The left side factors as (x + b/(2a))². The right side becomes −c/a + b²/(4a²). Write both terms over a common denominator 4a²: −c/a = −4ac/(4a²), so the right side is (b² − 4ac)/(4a²). Now take the square root of both sides. Remember the ±: x + b/(2a) = ± √(b² − 4ac) / (2a). Subtract b/(2a) from both sides to isolate x: x = −b/(2a) ± √(b² − 4ac) / (2a). Combine over the common denominator 2a to get x = [−b ± √(b² − 4ac)] / (2a).

The expression b² − 4ac under the radical is the discriminant. It determines the number and type of roots. If b² − 4ac > 0, there are two distinct real roots. If b² − 4ac = 0, there is exactly one real root (a repeated root). If b² − 4ac < 0, there are two complex conjugate roots.

The discriminant b² − 4ac emerges naturally at line 6 of the derivation. It is the term that must be non-negative for the square root to be real. This is not an extra concept bolted onto the quadratic formula; completing the square forces it into view.

Use the discriminant to check your work before plugging into the quadratic formula. Compute b² − 4ac first. If it is negative, stop: the quadratic formula will give complex numbers, and any real-world problem expecting a real answer is set up wrong. If it is zero, remember that the ± vanishes and the formula gives a single value.

The failure case most students encounter: they compute the discriminant correctly but then forget the denominator 2a when taking the square root, writing x = [−b ± √(b² − 4ac)] / a.

The completing the square formula and the quadratic formula are two sides of the same coin. After you complete the square to find the vertex (h, k), you can verify the vertex by plugging h into the original quadratic and seeing that f(h) = k. Alternatively, use the relationship h = −b/(2a) as a quick sanity check. If you solved a quadratic by completing the square and got a root, the quadratic formula should produce the same number.

Example: For f(x) = 2x² + 8x + 6, completing the square gives h = −2 and k = −2. Check that f(−2) = 2(4) + 8(−2) + 6 = 8 − 16 + 6 = −2, which matches k. Then solve 2x² + 8x + 6 = 0 by the quadratic formula: a = 2, b = 8, c = 6 gives x = [−8 ± √(64 − 48)] / 4 = [−8 ± √16] / 4 = [−8 ± 4] / 4. The roots are x = −1 and x = −3. Completing the square on the same equation should yield the same roots when you solve (x + 2)² = 1.

When the two methods disagree, the most common mistake is an arithmetic error in the completing the square process, especially the step where you factor a and add a·(b/(2a))² to the constant side. Use the quadratic formula as the tiebreaker because it has fewer steps and a single canonical form.