From Completed Square to the Square Root Property

You have completed the square and now hold an equation in the form (x − h)² = k. The next step is to apply the square root property, but you need to know when this step is valid and why the result carries a ± sign. The answer is immediate: you can take the square root of both sides of (x − h)² = k exactly when k is a real number, and you must retain both the positive and negative roots because the equation u² = d is always satisfied by both u = √d and u = −√d. The exact algebraic step where the bridge from completed square to the square root property occurs is walked through, one concrete example is worked, and the specific failure case that catches most students is named.

Work this example using inputs from the page fact sheet. Start with the equation (x − 3)² = 16. The left side is a perfect square, and the right side is a positive constant. The square root property applies here because k = 16 is a real number and 16 ≥ 0, so real solutions exist. Take the square root of both sides: x − 3 = ±√16. The result is x − 3 = ±4 because √16 = 4. Solve each branch: x = 3 + 4 = 7 and x = 3 − 4 = −1. Both satisfy the original equation because (7 − 3)² = 4² = 16 and (−1 − 3)² = (−4)² = 16. The two solutions are real and distinct, which matches the discriminant b² − 4ac = 0 for this equation (in its expanded form x² − 6x − 7 = 0, the discriminant is 36 + 28 = 64, a positive perfect square).

Now consider the same method when the squared term is not isolated. Suppose you have 2(x + 1)² − 8 = 0. Add 8 to both sides to get 2(x + 1)² = 8, then divide by 2 to isolate the squared term: (x + 1)² = 4. Apply the square root property: x + 1 = ±√4 = ±2, so x = −1 ± 2, giving x = 1 and x = −3. Every equation that can be rewritten as u² = d with d ≥ 0 yields two real solutions via this exact algebraic step.

The calculation itself is a two‑step operation after completing the square. First, verify that the squared term is alone on one side. Second, take the square root of both sides, always writing ± before the root. Use a square root calculator at this point to check the exact value of √k when k is not a perfect square. For example, if the completed square gives (x − 3)² = 5, then √5 is irrational. The exact solutions are x = 3 ± √5. Do not convert √5 to a decimal approximation unless the problem calls for it, because the exact form is required for later calculus work and for the discriminant analysis.

Why Both Signs Must Be Retained

The square root property is derived from the definition of square roots over the real numbers: for any non‑negative d, the equation u² = d has exactly two solutions because (−√d)² = (√d)² = d. If you drop the ±, you lose one solution. The supporting term both signs is not optional; it is a direct consequence of the field axioms of real numbers. On the page fact sheet, the equation (x − 3)² = 16 illustrates this: only writing x − 3 = 4 gives x = 7 and misses x = −1. That is a loss of 50% of the solution set.

The bridge from completed square to the square root property occurs at the exact algebraic step where you have isolated the perfect square trinomial. After completing the square on an equation like x² − 6x − 7 = 0, you rewrite it as (x − 3)² = 16. This step is the result of the half‑the‑coefficient rule: take b = −6, halve it to get −3, square it to get 9, add and subtract 9 inside the expression. The constant term c = −7 is then adjusted to produce k. The square root property applies immediately because the left side is a perfect square and the right side is a real number. The property holds for all real k, including k = 0 (one repeated real solution) and k < 0 (two complex solutions).

The Failure Case: k Is Negative

The most common failure occurs when the completed square yields a negative constant on the right. For example, from (x + 2)² = −9, the square root property still applies but extends to complex numbers: x + 2 = ±√(−9) = ±3i, so x = −2 ± 3i. Students who expect real solutions may believe the method has stopped working. It has not. The property is universal; the limitation is only that real solutions require k ≥ 0. The exact algebraic step is identical, but the result lives in the complex plane. The error rate for the add‑and‑subtract step during completion is the highest single mistake, but mishandling the ± sign when k is negative is the second most common failure.

The mistake that costs most students their grade is not the failure to retain the ± sign. It is the failure to isolate the squared term before applying the square root property. If you have 3(x − 4)² = 12, you must divide by 3 first to get (x − 4)² = 4. Applying the property to 3(x − 4)² = 12 directly by taking the square root of both sides gives √3 · (x − 4) = ±√12, which is messy and error‑prone. The correct order is always: isolate the squared term, then take the square root. The one thing you should do next is check that k is isolated and that you have written ± before taking the root. That single check prevents the loss of half your solutions.