Solve quadratic equations having real solutions by taking square roots and by completing the square (TEKS A.8A).

What this topic is asking

TEKS A.8A includes two more solving methods beyond factoring: taking square roots (for equations with no linear term, or already in squared form) and completing the square (which works on any quadratic and is the method behind vertex form and the quadratic formula). Both reliably appear in the Quadratic Functions and Equations category, often where factoring fails.

The square-root property

When a squared expression equals a number, take the square root of both sides, with both signs.

For $2x^2 = 50$: divide to get $x^2 = 25$, then $x = \pm 5$. For $(x + 1)^2 = 9$: $x + 1 = \pm 3$, so $x = 2$ or $x = -4$. The $\pm$ is essential; dropping it loses a solution.

Completing the square

Completing the square rewrites $x^2 + bx$ as a perfect square by adding $\left(\frac{b}{2}\right)^2$. It works on every quadratic and is required when a problem says "complete the square".

When the leading coefficient is not 1, divide every term by $a$ first so the $x^2$ coefficient is 1, then complete the square.

Simplest radical form

When the number under the root is not a perfect square, leave the answer in simplest radical form: pull out the largest perfect-square factor. Solving $(x - 1)^2 = 12$ gives $x - 1 = \pm\sqrt{12} = \pm 2\sqrt{3}$, so $x = 1 \pm 2\sqrt{3}$. A rounded decimal is not "exact" and can lose credit when exact form is requested, the same radical discipline as the quadratic formula.

How STAAR examines this topic

• Multiple choice. Solve a squared-form equation; the "positive root only" answer is the standard trap.
• Equation editor. Enter the solutions, often in simplest radical form, after completing the square.
• Connection to vertex form. Completing the square converts standard form to vertex form, linking solving to graphing.

A clarifying idea is that both methods rely on the same final step, undoing a square with a $\pm$ square root, so completing the square is really a way to manufacture the squared form that the square-root property then solves.

When to take square roots directly

The square-root method is the quickest of all when the quadratic has no linear term ($b = 0$) or already appears as a perfect square. Equations like $3x^2 - 27 = 0$ (isolate to $x^2 = 9$, so $x = \pm 3$) and $(x + 5)^2 = 49$ ($x + 5 = \pm 7$) are solved in a single step, with no factoring or formula needed. Spotting that a quadratic is missing its $bx$ term, or is given in squared form, signals that taking square roots is the intended fast route, and STAAR includes these to reward method selection.

Completing the square and vertex form

Completing the square does double duty: besides solving, it is the algebra that converts standard form to vertex form. Taking $f(x) = x^2 + 6x - 1$ and completing the square on $x^2 + 6x$ gives $f(x) = (x + 3)^2 - 10$, which reveals the vertex $(-3, -10)$ directly. This is why the technique matters even when a faster solving method exists: it is the bridge between the equation you can solve and the graph you can read, and STAAR sometimes asks you to rewrite a quadratic in vertex form by exactly this process.

Try this

Q1. Solve $x^2 = 49$. [1 point]

• Cue. $x = \pm 7$.

Q2. Solve $(x - 2)^2 = 5$ in simplest radical form. [1 point]

• Cue. $x - 2 = \pm\sqrt{5}$, so $x = 2 \pm \sqrt{5}$.