The Square Root Property After Completing the Square

You solve a squared expression like (x + 3)² = 4 and write x + 3 = ±2, but the reason for the ± sign is not an arbitrary convention. The square root property after completing the square demands it because every positive number has two real square roots: a principal (positive) root and a negative root. The definition from OpenStax Intermediate Algebra 2e states that if x² = k, then x = ±√k for k ≥ 0. This covers the exact question: Why does solving a squared expression require both positive and negative roots? Because squaring is a lossy operation, both 2 and -2 square to 4, so reversing it must recover both possibilities.

Work one concrete example from the page-specific fact sheet: x² + 6x + 5 = 0. Completing the square transforms this into (x + 3)² = 4. The squared expression here is (x + 3)², and the property states that if (x + 3)² = 4, then x + 3 = ±√4. Since √4 = 2, you get x + 3 = 2 or x + 3 = -2. Solving gives x = -1 or x = -5. Check both: (-1)² + 6(-1) + 5 = 1 - 6 + 5 = 0; (-5)² + 6(-5) + 5 = 25 - 30 + 5 = 0. Omitting the negative root means losing x = -5, which is a valid solution.

The calculation step uses the formula from the fact sheet: if (x + d)² = e and e ≥ 0, then x = -d ± √e. After you complete the square for x² + 6x + 5 = 0, you identify d = 3 and e = 4. The calculation gives x = -3 ± 2, producing two values. The completing the square calculator executes this same arithmetic when you enter the coefficients a = 1, b = 6, c = 5. It isolates the squared term, extracts the square root, and applies the ± symbol automatically. Understanding the calculation means seeing that the ± is not a decoration, it is the direct result of reversing the squaring operation.

For any real number a, a² = (-a)². This is a fundamental property of squaring: a negative base squared produces a positive result. When you complete the square to get (x + d)² = e, you are not solving an equation with a single answer, you are undoing the squaring. The principal square root √e gives one candidate, but because (-√e)² = e, the negative root is equally valid. OpenStax Intermediate Algebra 2e makes this explicit: the equation (x + d)² = e has two solutions, x + d = √e and x + d = -√e. The ± symbol is a shorthand for writing both branches.

Here is the limitation from the fact sheet: the square root property requires e ≥ 0. If completing the square produces a negative value for e, say (x + 3)² = -4, then the equation has no real solutions. The square root of a negative number is not a real number but an imaginary one, defined by the imaginary unit i where i = √(-1). In that case the solutions become complex: x = -3 ± 2i. The property still works, but the radicand enters the complex number system rather than the real numbers.

The most common failure when applying the square root property after completing the square is forgetting the negative root entirely. If you solve x² + 6x + 5 = 0 by completing the square and write only x + 3 = 2, you get x = -1 and stop. You lose x = -5. Checking the original equation shows x = -5 is a valid solution, so the omission is an error. The research under Failure Modes lists this as the first item: 'Forgetting the negative root, Solving x² = 9 and writing only x = 3, missing x = -3.' The exact same failure applies here. The fix is mechanical: always write the ± symbol after applying the square root, and solve both equations separately.

The √ symbol by convention denotes only the principal (non-negative) square root. When you read √4, the value is 2, not -2. This is the principal square root as defined in the research under Vocabulary: 'The non-negative square root of a non-negative number.' If the √ symbol already meant 'both', then writing ±√ would be redundant. But the research under Confusion Pairs is explicit: '√9 vs 3, √9 is the principal square root (3), but many think it is ±3; the ± appears only when solving x² = 9.' This distinction matters. When you write x + 3 = √4, you get x + 3 = 2. Only by adding the ± do you recover the negative branch: x + 3 = -2.

When e is not a perfect square, the same two-root requirement holds. Solve x² + 6x + 2 = 0 by completing the square. Step 1: x² + 6x = -2. Step 2: Add (6/2)² = 9: x² + 6x + 9 = 7. Step 3: (x + 3)² = 7. Step 4: x + 3 = ±√7. Step 5: x = -3 ± √7. Here √7 is an irrational number, not a simple integer. The two solutions are -3 + √7 and -3 - √7. Both are exact values. The ± is not a rounding choice, it is the only way to express both solutions exactly. A calculator that gives a decimal approximation for √7, like 2.645751311, does not change the requirement. The ± still means two separate numbers: approximately -0.354248689 and approximately -5.645751311.

Complete the square for x² + 6x + 13 = 0. Step 1: x² + 6x = -13. Step 2: Add 9: x² + 6x + 9 = -4. Step 3: (x + 3)² = -4. The value of e is -4, which is less than zero. The square root property as stated for real numbers requires e ≥ 0. Since -4 < 0, the equation has no real solutions. The square root of a negative number is defined in the complex number system using the imaginary unit i, where √(-4) = 2i. The property extends to x = -3 ± 2i, but this is a complex solution, not a real one. A standard calculator set to real-number mode returns an error for √(-4). This is the specific failure when the radicand is negative, the square root property after completing the square still works mathematically, but the output leaves the real numbers.

Completing the square is a method that rewrites any quadratic ax² + bx + c = 0 into the form (x + d)² = e. This is not an approximation, it is an algebraic identity. The transformation adds and subtracts the same value to create a perfect square trinomial. Once you have (x + d)² = e, the left side is exactly a squared expression. The square root property applies directly to that expression because it is of the form (something)² = a number. The property does not apply to the original quadratic in its standard form; it applies only after the completion step creates the squared binomial. That is why the two procedures are tied together: completing the square sets up the equation for the square root property to solve.

In the example x² + 6x + 5 = 0, the completed form is (x + 3)² = 4. The two solutions are x = -1 and x = -5. Their average is (-1 + -5)/2 = -3, which is exactly -d where d = 3. This is not a coincidence: for any equation (x + d)² = e, the two roots are -d + √e and -d - √e. Their midpoint is -d. The distance from the midpoint to each root is √e. This geometric property holds whether e is a perfect square, an irrational number, or even a negative value in the complex plane. The ± symbol encodes this symmetry, it says 'add the square root' and 'subtract the square root' from the same central value.

You complete the square correctly, get (x + 3)² = 4, and then write x + 3 = 2. You solve for x = -1 and stop. You miss the second root. The research under Failure Modes lists this as the first failure: 'Forgetting the negative root.' The fix is procedural: every time you see (something)² = a positive number, write the ± symbol before you write the square root. Do not write √4 first and then add the ±, write ±√4 as a single step. This forces you to handle both cases. If you are using a calculator, the SQRT function returns only the principal square root. You must supply the ± yourself by solving both x + 3 = √4 and x + 3 = -√4. The calculator does not know you need two answers.

After solving any quadratic by completing the square and applying the square root property, substitute both candidate solutions back into the original equation. This catches arithmetic errors and confirms that the ± truly covers both valid answers. For x² + 6x + 5 = 0, checking x = -1 and x = -5 takes less than a minute and verifies that neither root is extraneous. The single thing that most often goes wrong here is not the algebra, it is the assumption that one answer is enough. Every positive number has two real square roots, and every equation of the form (x + d)² = e with e > 0 has two real solutions. Write both.