Solve by Completing the Square

You have a quadratic equation like x² + 6x + 5 = 0 and factoring does not work, or the coefficients are messy. The direct method is to solve by completing the square. This rewrites the equation into the form (x − h)² = k, from which you extract roots by taking a square root. The method always works, for any quadratic, and it is the algebraic backbone behind the quadratic formula and vertex form.

To complete the square, you reverse-engineer a perfect square trinomial from the x² and x terms. The key step: take half the coefficient of x, square it, and add that number inside the parentheses, then adjust the constant to keep the equation balanced. The single most error-prone part is forgetting to factor out the coefficient a before adding inside when a ≠ 1. That mistake shifts the vertex's y-coordinate and gives wrong roots.

Below are three worked examples: rational roots, irrational roots, and no real roots (complex answers). Each follows the same step sequence. Use them as a template for any problem.

Start with the equation in standard form ax² + bx + c = 0. If a ≠ 1, factor a out of the first two terms. Move the constant term to the right side. Take half of the x-coefficient (inside the parentheses), square it, and add that value inside the parentheses. To keep the equation balanced, subtract a times that same square from the constant side. Factor the parentheses into a perfect square. Divide both sides by a (if a ≠ 1). Take the square root of both sides, remembering the ± sign. Solve for x.

Failure case: when b is odd, half of b is a fraction. Students freeze. The fix is to work with fractions exactly, do not convert to decimals. The half-the-coefficient rule still applies: (½b)².

Equation: x² + 6x + 5 = 0

Step 1: a = 1, so no factoring needed. Move the constant: x² + 6x = −5.

Step 2: Half of 6 is 3. Square it: 9. Add 9 inside the parentheses (since a = 1, you add 9 to both sides): x² + 6x + 9 = −5 + 9.

Step 3: Factor the left side: (x + 3)² = 4.

Step 4: Take the square root: x + 3 = ±2.

Step 5: Solve: x = −3 + 2 = −1, or x = −3 − 2 = −5.

Check: (−1)² + 6(−1) + 5 = 1 − 6 + 5 = 0. (−5)² + 6(−5) + 5 = 25 − 30 + 5 = 0.

Equation: 2x² − 4x − 6 = 0

Step 1: Factor a = 2 from the first two terms: 2(x² − 2x) − 6 = 0.

Step 2: Move the constant: 2(x² − 2x) = 6.

Step 3: Half of −2 is −1. Square it: 1. Add 1 inside the parentheses. Since a = 2, add 2·1 = 2 to the right side: 2(x² − 2x + 1) = 6 + 2 = 8.

Step 4: Factor the parentheses: 2(x − 1)² = 8.

Step 5: Divide by 2: (x − 1)² = 4.

Step 6: Take the square root: x − 1 = ±2.

Step 7: Solve: x = 1 + 2 = 3, or x = 1 − 2 = −1.

Check: 2(3)² − 4(3) − 6 = 18 − 12 − 6 = 0. 2(−1)² − 4(−1) − 6 = 2 + 4 − 6 = 0. Both rational, but the method handles irrational roots the same way.

Equation: x² + 2x + 5 = 0

Step 1: a = 1. Move the constant: x² + 2x = −5.

Step 2: Half of 2 is 1. Square it: 1. Add 1 to both sides: x² + 2x + 1 = −5 + 1 = −4.

Step 3: Factor: (x + 1)² = −4.

Step 4: Take the square root: x + 1 = ±√(−4). Since the radicand is negative, the roots are complex. Write √(−4) = 2i, where i = √(−1).

Step 5: Solve: x = −1 ± 2i.

Check: (−1 + 2i)² + 2(−1 + 2i) + 5 = (1 − 4i − 4) + (−2 + 4i) + 5 = (1 − 4 − 2 + 5) + (−4i + 4i) = 0 + 0 = 0.

The discriminant here is 2² − 4·1·5 = 4 − 20 = −16, negative, confirming no real solutions. The method gives complex roots directly.

Substitute each root back into the original equation. The result must be zero. Use exact arithmetic, if you approximated a radical, the check will not land on zero. For complex roots, verify that both the real and imaginary parts cancel.

If your answer involves a square root that does not simplify, leave it as an exact radical. Do not convert to a decimal. The decimal is an approximation and will not check exactly. Use a computer algebra system for heavy verification, but for homework, manual substitution catches sign errors.

The most common failure when checking: you expanded the square incorrectly because you forgot the factor a in the constant term. If the check fails, revisit the step where you added and subtracted inside the brackets.