Solve quadratic equations by taking square roots and by completing the square, including writing the equation in vertex form (MA.912.AR.3.4, MA.912.AR.3.8).

What this topic is asking

MA.912.AR.3 asks you to solve quadratics two more ways: the square-root property (for equations in squared form) and completing the square (which works on any quadratic and also produces vertex form). The B.E.S.T. Algebra 1 EOC tests both, often wanting answers in simplest radical form via the equation editor.

The square-root property

Apply it when the variable appears only inside a square (no separate linear term), or when the equation is already a squared binomial equal to a number. The $\pm$ produces the two solutions; a square root taken without it loses half the answer.

Completing the square

Completing the square forces a perfect-square trinomial so the square-root property applies.

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

Completing the square gives vertex form

The same steps, kept on one side, convert standard form to vertex form. For $f(x) = x^2 + 4x + 1$: $x^2 + 4x + 4 - 4 + 1 = (x + 2)^2 - 3$, so the vertex is $(-2, -3)$. This is why completing the square does double duty, solving and finding the vertex.

How the B.E.S.T. EOC examines this topic

• Equation editor. Solve in simplest radical form, or write a quadratic in vertex form.
• Multiple choice. Solve a squared-binomial equation, with "positive root only" distractors.
• Inline choice. Identify the value to add when completing the square.

A clarifying idea: completing the square is the manufacture of a perfect square where there was not one, by adding exactly the missing piece, $\left(\frac{b}{2}\right)^2$. Once the left side is a perfect square, the square-root property finishes the job, so completing the square is really a setup step for taking square roots.

Why you add (b/2) squared

The number you add is not a guess; it is the exact constant that completes the pattern of a perfect-square trinomial. A perfect square expands as $(x + m)^2 = x^2 + 2mx + m^2$, so the constant ($m^2$) is always the square of half the coefficient of $x$ (since the middle coefficient is $2m$, half of it is $m$, and squaring gives $m^2$). For $x^2 + 8x$, half of $8$ is $4$, and $4^2 = 16$ is the constant that makes $x^2 + 8x + 16 = (x + 4)^2$. You must add it to both sides to keep the equation balanced, which is the step students most often forget. Understanding the $(x + m)^2$ expansion means you can always recover the right number instead of memorizing the rule blindly.

When to take square roots versus completing the square

The two methods on this page suit different equations, and choosing well saves work. Take square roots directly when the quadratic has no linear term ($x^2 = 49$, so $x = \pm 7$) or is already a squared binomial equal to a number ($(x - 3)^2 = 10$). There is nothing to complete, just isolate the square and apply the property with the $\pm$. Reach for completing the square when there is a linear term and you either need an exact answer or want the vertex form along the way. Because completing the square always works, it is a reliable backup when a quadratic does not factor, though the quadratic formula (which is completing the square done once, in general) is usually quicker for a pure solve. A practical rule: if you see only $x^2$ and a constant, take roots; if you see $x^2$, $x$, and a constant and want the vertex, complete the square.

Simplest radical form

Equation-editor items often demand simplest radical form, and partial-credit scoring rewards the reduced answer. To simplify a radical, pull out the largest perfect-square factor: $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$, and $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$. When the radical sits in a solution like $x = \frac{-6 \pm \sqrt{20}}{2}$, simplify the radical first ($\sqrt{20} = 2\sqrt{5}$) and then reduce the whole fraction by the common factor: $\frac{-6 \pm 2\sqrt{5}}{2} = -3 \pm \sqrt{5}$. Leaving $\sqrt{20}$ unsimplified or forgetting to divide every term of the fraction by the common factor are the two errors that cost the point even when the method was right.

Try this

Q1. Solve $(x + 1)^2 = 25$. [1 point]

• Cue. $x + 1 = \pm 5$, so $x = 4$ or $x = -6$.

Q2. What do you add to complete the square for $x^2 + 10x$? [1 point]

• Cue. $\left(\frac{10}{2}\right)^2 = 25$.