Completar o Quadrado: Guia Passo a Passo com Exemplos Resolvidos
Completar o quadrado é uma técnica algébrica que reescreve uma expressão quadrática como um quadrado perfeito mais uma constante, tornando possível resolver equações que não podem ser fatoradas, converter forma padrão para forma de vértice e até derivar a fórmula quadrática. Aparece em álgebra do ensino médio, exames de admissão na universidade e cursos de cálculo sempre que expressões quadráticas surgem. Diferentemente da fórmula quadrática, que lhe dá uma resposta, completar o quadrado mostra como a resposta é construída — e essa compreensão compensa em muitos tópicos. Este guia aborda cada etapa com exemplos numéricos totalmente trabalhados, cobertura do caso mais difícil em que o coeficiente líder não é 1, derivação completa da fórmula quadrática e uma seção de FAQ que aborda as questões em que os alunos consistentemente ficam presos.
Conteúdo
- 01What Is Completing the Square?
- 02How to Complete the Square Step by Step (a = 1)
- 03Completing the Square When a ≠ 1
- 04Converting Standard Form to Vertex Form
- 05Deriving the Quadratic Formula by Completing the Square
- 06Common Mistakes When Completing the Square
- 07Practice Problems with Full Solutions
- 08When to Use This Method vs. Factoring or the Quadratic Formula
- 09FAQ — Completing the Square
What Is Completing the Square?
A quadratic expression in the form x² + bx + c does not automatically reveal its roots, its vertex, or its maximum and minimum value. Completing the square is the algebraic technique that reorganizes this expression into the form (x + p)² + q, where everything hidden in the standard form becomes visible at once. The key observation is that any perfect square binomial (x + p)² expands to x² + 2px + p². So if you start with x² + bx and want to create a perfect square trinomial, you need to add exactly (b/2)² — the square of half the coefficient of x. That added constant is what 'completes' the square. The resulting form is called vertex form when the technique is applied to a two-variable equation y = ax² + bx + c. After the conversion, the equation becomes y = a(x − h)² + k, where the vertex of the parabola is immediately visible as the point (h, k). When you are solving ax² + bx + c = 0 (setting the expression equal to zero), the technique rewrites the left side so that taking the square root of both sides is the obvious next step. Why learn this method when the quadratic formula exists? Three solid reasons. First, some problems — vertex form conversions, conic section equations, integration setups in calculus — specifically require this algebraic form rather than just the roots. Second, the quadratic formula itself is derived by completing the square on the general form ax² + bx + c = 0, so understanding the process gives you insight into where that formula comes from. Third, when the leading coefficient is 1 and the numbers are manageable, this approach is often faster than the formula. It belongs in your algebra toolkit alongside factoring and the quadratic formula — not instead of them.
Completing the square transforms x² + bx into a perfect square trinomial by adding (b/2)² to both sides. For y = ax² + bx + c, factor out a first, then add and subtract (b/(2a))² inside the parentheses. The result reveals the vertex of the parabola and converts the equation to vertex form y = a(x − h)² + k.
How to Complete the Square Step by Step (a = 1)
When the coefficient of x² is 1, the process follows a clean six-step sequence. All six steps are demonstrated below on x² + 6x + 1 = 0, then immediately repeated on a second example to confirm the pattern. Both equations have irrational solutions — the kind the quadratic formula handles but factoring cannot reach — which is exactly the situation where this method earns its place.
1. Step 1 — Move the constant to the right side
Rewrite the equation so the x² and x terms are on the left and the constant is on the right. For x² + 6x + 1 = 0, subtract 1 from both sides: x² + 6x = −1. If the constant is already 0 (for example, x² + 6x = 0), leave 0 on the right — the process still works identically.
2. Step 2 — Find the completing-the-square constant: (b/2)²
The coefficient of x is b = 6. Divide by 2 to get 3, then square: (6/2)² = 3² = 9. This is the number that, when added to x² + 6x, creates the perfect square trinomial x² + 6x + 9 = (x + 3)². Always square after dividing — do not just divide without squaring, and do not square before dividing.
3. Step 3 — Add the completing-the-square constant to both sides
Add 9 to both sides of the equation to keep equality: x² + 6x + 9 = −1 + 9, which gives x² + 6x + 9 = 8. The left side now contains the three terms of a perfect square trinomial. Adding to both sides preserves equality — this step is where many students add the constant to only one side and break the equation.
4. Step 4 — Factor the left side as a perfect square
The left side x² + 6x + 9 factors as (x + 3)². Write: (x + 3)² = 8. The number inside the parentheses is always b/2: here, 6/2 = 3. The rule is: x² + bx + (b/2)² always factors as (x + b/2)². No guessing is required.
5. Step 5 — Take the square root of both sides
Apply the square root to both sides: √[(x + 3)²] = ±√8. The left side simplifies to x + 3. The right side is ±√8 = ±2√2, because √8 = √(4 × 2) = 2√2. Write: x + 3 = ±2√2. The ± sign is not optional — one root comes from the positive square root and one from the negative, and omitting ± loses one solution entirely.
6. Step 6 — Solve for x
Subtract 3 from both sides: x = −3 ± 2√2. This gives two solutions: x = −3 + 2√2 ≈ −0.17 and x = −3 − 2√2 ≈ −5.83. Check by substituting x = −3 + 2√2 into the original equation: x² + 6x + 1 = (−3 + 2√2)² + 6(−3 + 2√2) + 1 = (9 − 12√2 + 8) + (−18 + 12√2) + 1 = 17 − 12√2 − 18 + 12√2 + 1 = 0 ✓.
7. Worked Example 2 — x² − 8x + 3 = 0
Step 1: x² − 8x = −3. Step 2: b = −8; constant = (−8/2)² = (−4)² = 16. The square of a negative is positive, so the constant is always non-negative. Step 3: x² − 8x + 16 = −3 + 16 = 13. Step 4: (x − 4)² = 13. The sign inside is b/2 = −4: write (x − 4), not (x + 4). Step 5: x − 4 = ±√13. Step 6: x = 4 ± √13. Numerically: x ≈ 7.61 or x ≈ 0.39. Vieta's check: sum of roots = 4 + √13 + 4 − √13 = 8 = −(−8)/1 = −b/a ✓.
For x² + bx: the constant to add is (b/2)². Add it to both sides, factor the left as (x + b/2)², then take the square root and solve. The ± on the square root is mandatory — it produces both solutions.
Completing the Square When a ≠ 1
When the coefficient of x² is not 1, one extra step comes first: factor out the leading coefficient from the x² and x terms. The constant c is left outside. This brings the expression inside the parentheses to the form x² + (b/a)x — a leading coefficient of 1 — where the standard method applies. The critical detail is that when the completing-the-square constant is added inside the parentheses, it is multiplied by a when moved outside, which changes the arithmetic on the right side.
1. Worked Example 1 — 2x² − 12x + 5 = 0
Step 1: Move the constant: 2x² − 12x = −5. Step 2: Factor out a = 2 from the left side: 2(x² − 6x) = −5. Step 3: Find the constant for the expression inside. Coefficient of x inside is −6; constant = (−6/2)² = (−3)² = 9. Step 4: Add 9 inside the parentheses. Because 9 is inside parentheses multiplied by 2, adding 9 inside adds 2 × 9 = 18 to the left side. Add 18 to the right side: 2(x² − 6x + 9) = −5 + 18 = 13. Step 5: Factor the perfect square trinomial: 2(x − 3)² = 13. Step 6: Divide both sides by 2: (x − 3)² = 13/2. Step 7: x − 3 = ±√(13/2) = ±√26/2. Step 8: x = 3 ± √26/2. Numerically: √26 ≈ 5.099, so x ≈ 5.55 or x ≈ 0.45.
2. Worked Example 2 — 3x² + 6x − 2 = 0
Step 1: 3x² + 6x = 2. Step 2: Factor out 3: 3(x² + 2x) = 2. Step 3: Constant = (2/2)² = 1² = 1. Adding 1 inside adds 3 × 1 = 3 to the left side; add 3 to the right: 3(x² + 2x + 1) = 2 + 3 = 5. Step 4: 3(x + 1)² = 5. Step 5: (x + 1)² = 5/3. Step 6: x + 1 = ±√(5/3) = ±√15/3. Step 7: x = −1 ± √15/3. Numerically: √15 ≈ 3.873, so x ≈ 0.291 or x ≈ −2.291. Verify with the quadratic formula: x = (−6 ± √(36 + 24)) / 6 = (−6 ± √60) / 6 = (−6 ± 2√15) / 6 = −1 ± √15/3 ✓.
3. Alternative: Divide Through by a First
Some teachers prefer dividing the entire equation by a before proceeding, eliminating the leading coefficient immediately. For 2x² − 12x + 5 = 0, divide by 2: x² − 6x + 5/2 = 0. Move 5/2 to the right: x² − 6x = −5/2. Add (−6/2)² = 9: x² − 6x + 9 = −5/2 + 9 = 13/2. Factor: (x − 3)² = 13/2. This gives the same result. The trade-off: fractions appear earlier, but you avoid tracking the factor of a through the rest of the computation. Either approach is correct.
When a ≠ 1: factor out a from the x² and x terms, leaving c outside. Complete the square inside the parentheses. Remember that the constant added inside is multiplied by a when moved outside — compensate by adding a × (b/2a)² to the right side, not just (b/2a)².
Converting Standard Form to Vertex Form
One of the most practical applications of this technique is converting y = ax² + bx + c into vertex form y = a(x − h)² + k. Vertex form immediately shows the vertex (h, k), the axis of symmetry x = h, and the direction the parabola opens. This conversion is required in problems that ask you to graph a parabola, identify its maximum or minimum, or write the equation given a vertex. The process is nearly identical to solving by completing the square, with one key difference: because you are working with an equation in two variables, you do not move c to the other side. Instead, you add and subtract the same constant on one side, so the equation stays balanced without rearranging.
1. Worked Example 1 — Convert y = 2x² − 8x + 5 to vertex form
Step 1: Group the x² and x terms: y = (2x² − 8x) + 5. Step 2: Factor out a = 2: y = 2(x² − 4x) + 5. Step 3: Constant = (−4/2)² = (−2)² = 4. Step 4: Add and subtract 4 inside the parentheses: y = 2(x² − 4x + 4 − 4) + 5. Step 5: Separate the perfect square from the −4: y = 2(x² − 4x + 4) + 2(−4) + 5. The −4 exits the parentheses multiplied by 2. Step 6: Simplify: y = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3. Vertex form: y = 2(x − 2)² − 3. Vertex: (2, −3). Parabola opens upward (a = 2 > 0), minimum at (2, −3). Axis of symmetry: x = 2. Cross-check: h = −(−8)/(2 × 2) = 8/4 = 2 ✓; k = 2(4) − 8(2) + 5 = 8 − 16 + 5 = −3 ✓.
2. Worked Example 2 — Convert y = −x² + 6x − 4 to vertex form
Step 1: Group: y = (−x² + 6x) − 4. Step 2: Factor out a = −1: y = −(x² − 6x) − 4. Step 3: Constant = (−6/2)² = (−3)² = 9. Step 4: Add and subtract 9 inside: y = −(x² − 6x + 9 − 9) − 4. Step 5: y = −(x² − 6x + 9) − (−1)(9) − 4 = −(x − 3)² + 9 − 4. Vertex form: y = −(x − 3)² + 5. Vertex: (3, 5). Parabola opens downward (a = −1 < 0), maximum at (3, 5). The function value can never exceed 5. Range: y ≤ 5.
To convert y = ax² + bx + c to vertex form: factor out a from the x terms, add and subtract (b/(2a))² inside the parentheses (do NOT move it to the other side), simplify. The vertex (h, k) appears directly in y = a(x − h)² + k.
Deriving the Quadratic Formula by Completing the Square
Every time you use x = (−b ± √(b² − 4ac)) / (2a), you are using a result that was derived by applying this algebraic technique to the general form ax² + bx + c = 0. Understanding the derivation is worth the effort: it shows that the formula is not arbitrary, it deepens your feel for the mechanics in the hardest possible case (general a, b, c), and it gives you something to reconstruct if you ever forget the formula on an exam. The five steps below follow the same sequence used in every specific numerical example above.
1. Step 1 — Move c to the right side
Start with ax² + bx + c = 0. Subtract c from both sides: ax² + bx = −c.
2. Step 2 — Divide every term by a
Divide through by a (valid because a ≠ 0 for any quadratic): x² + (b/a)x = −c/a. Now the leading coefficient is 1 and the standard process can continue.
3. Step 3 — Find and add the completing-the-square constant
The coefficient of x is b/a. Half of that is b/(2a). Square it: [b/(2a)]² = b²/(4a²). Add this to both sides: x² + (b/a)x + b²/(4a²) = −c/a + b²/(4a²).
4. Step 4 — Factor the left side and simplify the right side
The left side is a perfect square: (x + b/(2a))² = b²/(4a²) − c/a. Combine the right side over the common denominator 4a²: rewrite −c/a as −4ac/(4a²). The right side becomes (b² − 4ac)/(4a²). This is the discriminant in the numerator.
5. Step 5 — Take the square root and isolate x
Take the square root of both sides: x + b/(2a) = ±√[(b² − 4ac)/(4a²)] = ±√(b² − 4ac) / (2a). Subtract b/(2a) from both sides: x = −b/(2a) ± √(b² − 4ac)/(2a) = [−b ± √(b² − 4ac)] / (2a). This is the quadratic formula. Every term in it came directly from completing the square on the general form — the discriminant b² − 4ac is the quantity remaining after forming the perfect square on the left side.
The quadratic formula x = (−b ± √(b² − 4ac)) / (2a) is the result of completing the square on ax² + bx + c = 0 in full generality. The discriminant b² − 4ac appears because it is the number left on the right side after the left side becomes a perfect square.
Common Mistakes When Completing the Square
Students learning this technique make several predictable errors. Each one below is paired with its source and the correct approach. Reviewing this list after your first practice session is a reliable way to catch habits before they become ingrained — most of these mistakes cost a mark on exams without the student realizing what went wrong.
1. Mistake 1 — Adding the constant to only one side
The most common error: adding (b/2)² to the left side but not the right. For x² + 6x = −1, you must add 9 to both sides: x² + 6x + 9 = −1 + 9 = 8. Writing x² + 6x + 9 = −1 breaks the equation — the two sides are no longer equal. Every number added to one side must be added to the other.
2. Mistake 2 — Squaring b instead of b/2
The constant to add is (b/2)², not b². For x² + 10x: the constant is (10/2)² = 5² = 25, not 10² = 100. A useful mental check: ask what binomial squares to give x² + 10x + ?: the answer is (x + 5)² = x² + 10x + 25, so the constant is 25. The number inside the binomial is always b/2, not b.
3. Mistake 3 — Forgetting the factor of a when moving the constant outside
When a ≠ 1 and you add a constant inside the parentheses, that constant is multiplied by a when it exits. For 3(x² + 4x + 4 − 4): the −4 exits multiplied by 3, giving 3(x + 2)² − 12. A student who writes 3(x + 2)² − 4 is off by 2 × 4 = 8. Write out 3(x + 2)² + 3(−4) explicitly before simplifying to avoid this.
4. Mistake 4 — Wrong sign inside the factored binomial
After factoring the perfect square trinomial, the number inside the parentheses is b/2, not b. For x² − 8x + 16, the factored form is (x − 4)², not (x − 8)². The rule: x² + bx + (b/2)² = (x + b/2)². When b is negative, b/2 is also negative: for b = −8, b/2 = −4, so the factor is (x + (−4)) = (x − 4).
5. Mistake 5 — Omitting the ± when taking the square root
When you write √[(x − 4)²] = √13, the result is x − 4 = ±√13, not x − 4 = √13. Every positive real number has two square roots. Omitting ± always discards one solution. In exam problems that ask for 'all solutions' or 'how many real roots', this error directly causes a wrong answer.
6. Mistake 6 — Leaving the square root unsimplified
If the right side is √8, simplify it: √8 = √(4 × 2) = 2√2. Leaving x = −3 ± √8 is technically correct but is not in simplest radical form, and many grading rubrics require simplification. After taking the square root, factor out the largest perfect square from under the radical: look for factors of 4, 9, 16, 25, and so on.
Practice Problems with Full Solutions
Work through each problem independently before reading the solution. Problems 1 and 2 have a leading coefficient of 1 and clean integers. Problem 3 has a common factor that simplifies things once you divide it out. Problem 4 has a ≠ 1 with no common factor. Problem 5 asks for vertex form and additional features of the parabola.
1. Problem 1 (Easy) — Solve x² + 4x − 3 = 0
Step 1: x² + 4x = 3. Step 2: (4/2)² = 4. Step 3: x² + 4x + 4 = 3 + 4 = 7. Step 4: (x + 2)² = 7. Step 5: x + 2 = ±√7. Step 6: x = −2 ± √7. Solutions: x = −2 + √7 ≈ 0.646 and x = −2 − √7 ≈ −4.646. Verify the positive root: (−2 + √7)² + 4(−2 + √7) − 3 = (4 − 4√7 + 7) + (−8 + 4√7) − 3 = 11 − 4√7 − 8 + 4√7 − 3 = 0 ✓.
2. Problem 2 (Easy) — Solve x² − 10x + 20 = 0
Step 1: x² − 10x = −20. Step 2: (−10/2)² = 25. Step 3: x² − 10x + 25 = −20 + 25 = 5. Step 4: (x − 5)² = 5. Step 5: x − 5 = ±√5. Step 6: x = 5 ± √5. Solutions: x = 5 + √5 ≈ 7.236 and x = 5 − √5 ≈ 2.764. Vieta's check: sum of roots = (5 + √5) + (5 − √5) = 10 = −(−10)/1 ✓. Product of roots = (5 + √5)(5 − √5) = 25 − 5 = 20 = c/a ✓.
3. Problem 3 (Medium) — Solve 2x² + 4x − 6 = 0
Spot that all coefficients share a factor of 2. Divide through by 2 first: x² + 2x − 3 = 0. Now a = 1 and the numbers are small. Step 1: x² + 2x = 3. Step 2: (2/2)² = 1. Step 3: x² + 2x + 1 = 4. Step 4: (x + 1)² = 4. Step 5: x + 1 = ±2. Step 6: x = −1 ± 2. Solutions: x = 1 or x = −3. Confirm by factoring the divided equation: (x − 1)(x + 3) = 0 ✓. When a shares a factor with b and c, always divide first — it avoids working with fractions.
4. Problem 4 (Medium) — Solve 4x² − 24x + 11 = 0
No common factor among 4, 24, 11. Use the standard a ≠ 1 procedure. Step 1: 4x² − 24x = −11. Step 2: Factor out 4: 4(x² − 6x) = −11. Step 3: Constant = (−6/2)² = 9. Adding 9 inside adds 4 × 9 = 36 to the left side; add 36 to the right: 4(x² − 6x + 9) = −11 + 36 = 25. Step 4: 4(x − 3)² = 25. Step 5: (x − 3)² = 25/4. Step 6: x − 3 = ±5/2. Step 7: x = 3 ± 5/2. Solutions: x = 3 + 5/2 = 11/2 and x = 3 − 5/2 = 1/2. Verify by factoring: 4x² − 24x + 11 = (2x − 11)(2x − 1) → x = 11/2 or x = 1/2 ✓.
5. Problem 5 (Hard) — Convert y = 3x² + 12x − 1 to vertex form; state the vertex, axis of symmetry, and direction of opening
Step 1: Group: y = (3x² + 12x) − 1. Step 2: Factor out 3: y = 3(x² + 4x) − 1. Step 3: (4/2)² = 4. Step 4: Add and subtract 4 inside: y = 3(x² + 4x + 4 − 4) − 1. Step 5: y = 3(x² + 4x + 4) + 3(−4) − 1 = 3(x + 2)² − 12 − 1. Step 6: y = 3(x + 2)² − 13. Vertex form: y = 3(x + 2)² − 13. Note: (x + 2) = (x − (−2)), so h = −2 and k = −13. Vertex: (−2, −13). Axis of symmetry: x = −2. Direction: opens upward (a = 3 > 0), minimum at (−2, −13). Cross-check with the vertex formula: h = −12/(2 × 3) = −12/6 = −2 ✓; k = 3(4) + 12(−2) − 1 = 12 − 24 − 1 = −13 ✓.
When to Use This Method vs. Factoring or the Quadratic Formula
Completing the square is not always the fastest approach. Knowing when to use it — and when another method is quicker — saves time on timed tests and reduces arithmetic errors. Factoring is fastest when the equation has small integer coefficients and the discriminant (b² − 4ac) is a perfect square. For x² + 5x + 6 = 0, spotting (x + 2)(x + 3) = 0 takes ten seconds. Running through the six-step procedure would produce the same answer more slowly. Completing the square is the right choice in three specific situations: (1) the problem explicitly asks for vertex form, not just the roots; (2) the leading coefficient is 1 and the x-coefficient is even, giving a clean integer for (b/2)²; (3) the expression appears inside a conic section or integral where the squared form is the end goal. The quadratic formula works for every quadratic with no exceptions, but it involves the most arithmetic, especially when a, b, or c are large. If you are ever uncertain and time is limited, the formula will always get you to the answer. For most standard-form equations on algebra exams, though, it is worth scanning for factoring first, checking whether a = 1 and b is even (favors completing the square), and falling back to the formula only if neither method fits cleanly.
Use factoring when coefficients are small integers and the trinomial factors in a few seconds. Use completing the square when vertex form is needed or when a = 1 and b is even. Use the quadratic formula when the discriminant is not a perfect square or the coefficients are large and messy.
FAQ — Completing the Square
These are the questions students ask most often about this topic. The answers focus on the mechanical details that cause confusion and on how the method connects to other algebra topics.
1. What is completing the square used for?
The technique has three main uses: (1) solving quadratic equations that cannot be factored — equations with irrational or complex roots; (2) converting y = ax² + bx + c to vertex form y = a(x − h)² + k, which shows the vertex, axis of symmetry, and maximum or minimum directly; and (3) deriving the quadratic formula — which is just the result of applying the technique to ax² + bx + c = 0 in full generality with a, b, c as symbols.
2. How do you know what number to add when completing the square?
The number to add is always (b/2)², where b is the coefficient of x once the x² term has coefficient 1. Divide the x-coefficient by 2, then square that result. For x² + 10x: b = 10; add (10/2)² = 25. For x² − 7x: b = −7; add (−7/2)² = 49/4. The constant is always positive because you are squaring. If a ≠ 1, factor out a first so the coefficient of x² inside the parentheses is 1.
3. Can you complete the square when a is negative?
Yes. Factor out a (which is negative) from the x² and x terms, leaving a coefficient of 1 on x² inside the parentheses. For y = −2x² + 8x − 3: factor out −2 to get y = −2(x² − 4x) − 3. Complete the square inside: (−4/2)² = 4. Add and subtract 4 inside: y = −2(x² − 4x + 4 − 4) − 3 = −2(x − 2)² + 8 − 3 = −2(x − 2)² + 5. Vertex: (2, 5), parabola opens downward.
4. What happens when the right side is negative after completing the square?
A negative right side means the equation has no real solutions — the discriminant is negative. For x² + 2x + 5 = 0: x² + 2x = −5; add 1: (x + 1)² = −4. Since no real number squared gives a negative result, there are no real roots. In the complex number system, √(−4) = 2i, giving x = −1 ± 2i. But for a standard algebra course, a negative right side means no real solutions.
5. Is completing the square the same as the quadratic formula?
They are related but not identical. The quadratic formula is derived by applying completing the square to the general form ax² + bx + c = 0 with symbolic coefficients (see the derivation section above). Once the formula is derived, it is a shortcut: plug in a, b, c without repeating the full process. Completing the square is more flexible — it can produce vertex form rather than just roots — while the formula only gives roots.
6. Does completing the square work when b is odd?
Yes, though it introduces fractions. For x² + 5x + 3 = 0: b = 5; constant = (5/2)² = 25/4. Move 3 to the right: x² + 5x = −3. Add 25/4 to both sides: x² + 5x + 25/4 = −3 + 25/4 = −12/4 + 25/4 = 13/4. Factor: (x + 5/2)² = 13/4. Take the square root: x + 5/2 = ±√13/2. Solve: x = (−5 ± √13)/2. The fractions are unavoidable when b is odd, but the procedure is unchanged.
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