Completing the Square — Step-by-Step Guide
Completing the square is a technique for rewriting a quadratic expression ax² + bx + c in the form a(x − h)² + k, called vertex form. The vertex of the parabola is immediately visible as (h, k), and the technique is the basis for the derivation of the quadratic formula itself.
Unlike factoring (which only works for certain trinomials) and the quadratic formula (which is mechanical), completing the square reveals the structure of a quadratic — making it indispensable for graphing, optimization, and integration in calculus.
Formula
ax²+bx+c → a(x + b/2a)² + (c − b²/4a) | Vertex: h = −b/2a, k = c − b²/4a
The 5-Step Method for Completing the Square
Follow these steps for any quadratic ax² + bx + c:
Step 1: Factor out a from the first two terms (if a ≠ 1): a(x² + (b/a)x) + c.
Step 2: Find the "completing number" = (b/2a)². This is the constant that makes x² + (b/a)x + (b/2a)² a perfect square.
Step 3: Add and subtract that number inside the bracket: a(x² + (b/a)x + (b/2a)² − (b/2a)²) + c.
Step 4: Write the first three terms as a perfect square: a((x + b/2a)² − (b/2a)²) + c.
Step 5: Expand and simplify: a(x + b/2a)² − b²/4a + c = a(x + b/2a)² + (c − b²/4a).
The vertex is at h = −b/2a and k = c − b²/4a.
Example 1: Monic Quadratic (a = 1)
Convert f(x) = x² + 6x + 5 to vertex form.
Step 1: a = 1, so no initial factoring needed.
Step 2: Completing number = (6/2)² = 9.
Step 3: Write x² + 6x + 9 − 9 + 5.
Step 4: (x + 3)² − 9 + 5.
Step 5: f(x) = (x + 3)² − 4.
Vertex: (−3, −4). The parabola has its minimum at x = −3 where f(x) = −4. Roots: set (x+3)² = 4 → x+3 = ±2 → x = −1 or x = −5.
Example 2: Non-Monic Quadratic (a ≠ 1)
Convert f(x) = 2x² − 8x + 3 to vertex form.
Step 1: Factor 2 from first two terms: 2(x² − 4x) + 3.
Step 2: Completing number inside the bracket = (−4/2)² = (−2)² = 4.
Step 3: Add and subtract 4 inside: 2(x² − 4x + 4 − 4) + 3.
Step 4: 2((x − 2)² − 4) + 3.
Step 5: 2(x − 2)² − 8 + 3 = 2(x − 2)² − 5.
Vertex: (2, −5). Since a = 2 > 0, this is the minimum. Verify: f(2) = 2(4) − 16 + 3 = −5 ✓.
Example 3: Negative Leading Coefficient
Convert f(x) = −3x² + 12x − 7 to vertex form.
Step 1: Factor −3 from first two terms: −3(x² − 4x) − 7.
Step 2: Completing number = (−4/2)² = 4.
Step 3: −3(x² − 4x + 4 − 4) − 7.
Step 4: −3((x−2)² − 4) − 7.
Step 5: −3(x−2)² + 12 − 7 = −3(x−2)² + 5.
Vertex: (2, 5). Since a = −3 < 0, this is the maximum. The parabola opens downward. Maximum value is 5, occurring at x = 2.
Solving Quadratic Equations by Completing the Square
Completing the square can solve quadratic equations as an alternative to the quadratic formula.
Example: Solve x² + 4x − 12 = 0.
Complete the square: x² + 4x + 4 − 4 − 12 = 0 → (x+2)² = 16 → x+2 = ±4 → x = 2 or x = −6.
Advantage: The method shows you the vertex and both roots simultaneously. The roots are x = h ± √(k is the vertex value)... more precisely, roots are where a(x−h)² + k = 0 → (x−h)² = −k/a → x = h ± √(−k/a).
When is this better than the formula? When a = 1 and b is even — the arithmetic is cleaner. For example x² + 10x − 11 = 0: completing gives (x+5)² = 36 → x = 1 or x = −11, faster than applying the formula.
Completing the Square Derives the Quadratic Formula
The quadratic formula is simply completing the square applied to the general equation ax² + bx + c = 0:
Step 1: Divide by a: x² + (b/a)x + c/a = 0.
Step 2: Complete the square: x² + (b/a)x + (b/2a)² = (b/2a)² − c/a.
Step 3: (x + b/2a)² = (b² − 4ac) / 4a².
Step 4: x + b/2a = ±√(b² − 4ac) / 2a.
Step 5: x = (−b ± √(b² − 4ac)) / 2a. This is the quadratic formula.
Every time you use the quadratic formula, you are implicitly completing the square. Understanding this connection makes both techniques more intuitive.
Finding the Vertex by Completing the Square
The vertex of a parabola f(x) = ax² + bx + c is found by completing the square to get vertex form a(x − h)² + k. The vertex is (h, k).
Quick formula: h = −b/(2a) and k = f(h) = c − b²/(4a). If you only need the vertex, plug h back into the original equation.
Example: Find the vertex of y = x² − 4x + 1
Complete the square: x² − 4x + 4 − 4 + 1 = (x − 2)² − 3. Vertex: (2, −3).
Shortcut: h = −(−4)/(2·1) = 2. k = f(2) = 4 − 8 + 1 = −3. Vertex: (2, −3) ✓
Example: Find the vertex of y = 2x² + 8x + 5
h = −8/(2·2) = −2. k = f(−2) = 2(4) + 8(−2) + 5 = 8 − 16 + 5 = −3. Vertex: (−2, −3).
Verify via completing the square: 2(x² + 4x) + 5 = 2(x² + 4x + 4 − 4) + 5 = 2(x+2)² − 8 + 5 = 2(x+2)² − 3. Vertex: (−2, −3) ✓
Example: Find the vertex of y = −x² + 6x − 2
h = −6/(2·(−1)) = 3. k = f(3) = −9 + 18 − 2 = 7. Vertex: (3, 7). Since a < 0, this is the maximum.
Vertex form → standard form: Expand a(x − h)² + k = ax² − 2ahx + ah² + k. Compare coefficients: b = −2ah, c = ah² + k.
Applications: Optimization and Graphing
Maximum/minimum value: Vertex form a(x−h)²+k makes the maximum/minimum immediately visible. If a > 0, minimum value is k at x = h. If a < 0, maximum value is k at x = h.
Example — fencing problem: A farmer has 40 m of fencing to enclose a rectangular area against a barn wall. Maximize area A = x(40−2x) = −2x² + 40x. Complete the square: −2(x²−20x) = −2(x²−20x+100−100) = −2(x−10)² + 200. Maximum area = 200 m² at x = 10 m.
Revenue/profit: Profit P(x) = −2x² + 80x − 600 (x = units in hundreds). Vertex: h = 80/4 = 20 hundreds = 2000 units. k = −2(400) + 80(20) − 600 = −800 + 1600 − 600 = 200. Maximum profit = $200K at x = 2000 units.
Graphing parabolas: Vertex form tells you: (1) vertex location, (2) direction of opening (sign of a), (3) whether parabola crosses x-axis (compare k and a). All from one compact expression.