Solve quadratic equations by the square-root property and by completing the square, and use completing the square to rewrite a quadratic in vertex form (A.PAR, Patterning and Algebraic Reasoning).

What this topic is asking

This Patterning and Algebraic Reasoning (A.PAR) standard covers two related solving methods: the square-root property (for quadratics in squared form or with no linear term) and completing the square (which works on any quadratic and also produces vertex form). The Georgia Milestones EOC tests these as numeric-entry items asking for solutions in simplest radical form, and completing the square is the conceptual link to the transformations topic, because it converts standard form into the vertex form $a(x - h)^2 + k$. The recurring discipline is the $\pm$ when taking a square root.

The square-root property

When a quadratic is already a perfect square equal to a number, take the square root of both sides, remembering both signs.

For $(x - 2)^2 = 49$: $x - 2 = \pm 7$, so $x = 9$ or $x = -5$. The $\pm$ is essential; without it you lose one solution. This method also handles equations like $x^2 = 20$ (no linear term): $x = \pm\sqrt{20} = \pm 2\sqrt{5}$.

Completing the square

Completing the square rewrites any quadratic so the square-root property applies. The key step adds the square of half the linear coefficient.

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

Completing the square gives vertex form

The same procedure, applied to a function rather than an equation, converts standard form to vertex form. For $f(x) = x^2 + 6x + 2$: $f(x) = (x^2 + 6x + 9) - 9 + 2 = (x + 3)^2 - 7$, which is vertex form with vertex $(-3, -7)$. So completing the square both solves equations and reveals the vertex, tying this topic to transformations.

How the Milestones examines this topic

• Numeric entry. Solve by the square-root property or completing the square in simplest radical form.
• Multiple choice. Choose the solutions of a squared equation, with the "forgot the $\pm$" distractor.
• Constructed response. Complete the square to solve, or to rewrite a function in vertex form and state the vertex.

Why adding (b/2) squared works

The reason you add exactly $\left(\frac{b}{2}\right)^2$ is that this is the precise constant that makes $x^2 + bx + \square$ a perfect square. Expanding $(x + p)^2$ gives $x^2 + 2px + p^2$, so to match the middle term $bx$ you need $2p = b$, that is $p = \frac{b}{2}$, and then the constant must be $p^2 = \left(\frac{b}{2}\right)^2$. That is why "half the linear coefficient, then square it" is the recipe: half undoes the doubling in $2p$, and squaring produces the matching constant. Seeing this means you never have to guess the number to add, and it explains why completing the square always works regardless of whether the quadratic factors nicely, which is its main advantage over factoring.

Choosing between methods

The three solving methods (factoring, square roots and completing the square, the quadratic formula) each fit certain quadratics. Factor first if integer factors exist; it is fastest. Use the square-root property when there is no linear term or the equation is already squared. Complete the square when asked, or when you also want the vertex. Use the quadratic formula when nothing factors. On the EOC, recognizing which method the quadratic invites turns a slow problem into a quick one, and completing the square is the one to reach for whenever the question also involves the vertex or vertex form.

Try this

Q1. Solve $(x + 1)^2 = 16$. [1 point]

• Cue. $x + 1 = \pm 4$, so $x = 3$ or $x = -5$.

Q2. Solve $x^2 - 4x - 3 = 0$ by completing the square. [2 points]

• Cue. $x^2 - 4x = 3$; add 4: $(x - 2)^2 = 7$; $x = 2 \pm \sqrt{7}$.