Solve quadratic equations of the form x^2 = k or a(x - h)^2 = k by taking square roots, including the plus-or-minus sign, and recognize when there is no real solution (A.EI.6).

What this topic is asking

This part of A.EI.6 asks you to solve a quadratic by taking square roots, the fastest method when the equation has a squared term and no linear ($bx$) term. On the Virginia Algebra I SOL these are Equations and Inequalities items: solve $x^2 = k$ or $a(x - h)^2 = k$ and type the solutions, often as exact radicals. They appear as fill-in-the-blank and multiple choice.

When to take square roots

The square-root method is the right tool when the equation can be written as (something squared) equals a number, with no separate linear term. Examples: $x^2 = 25$, $3x^2 = 48$, $(x - 4)^2 = 9$. If there is an $x$ term that will not go away (like $x^2 + 6x = 7$), use factoring, completing the square, or the quadratic formula instead.

The plus-or-minus sign

Taking the square root to solve introduces two answers, because squaring erases sign: $(+7)^2 = 49$ and $(-7)^2 = 49$. So $x^2 = 49$ has solutions $x = +7$ and $x = -7$, written compactly as $x = \pm 7$. The $\pm$ is not optional, omitting it loses a solution, which is the single most common error here.

Squared binomials and radical answers

When the square is a binomial, square-root both sides and then solve the resulting linear equation. For $(x - 5)^2 = 16$: take roots, $x - 5 = \pm 4$, so $x = 5 \pm 4$, giving $x = 9$ or $x = 1$.

Sometimes $k$ is not a perfect square, and the answer stays as a simplified radical: $x^2 = 12$ gives $x = \pm\sqrt{12} = \pm 2\sqrt{3}$. Simplify the radical fully, just as in the radicals topic.

No real solution

If isolating the square leaves it equal to a negative number, the equation has no real solution. For $x^2 = -9$, no real number squares to $-9$ (a square is always $\ge 0$), so there is no real $x$. This is the same idea you will see in the discriminant: a negative under the square root means no real roots.

Why square-rooting needs the plus-or-minus

The reason solving by square roots produces two answers is that squaring is a two-to-one operation: it sends both $a$ and $-a$ to the same output $a^2$. So when you run it backward by taking a square root to solve an equation, you must recover both inputs that could have produced the value, which is exactly what $\pm$ records. The symbol $\sqrt{49}$ by itself denotes the single principal (non-negative) root, $7$. But the equation $x^2 = 49$ asks for every number whose square is $49$, and there are two. Keeping the principal-root symbol and the equation-solving step distinct explains why $\sqrt{49} = 7$ (one value) while the solutions of $x^2 = 49$ are $\pm 7$ (two values). The same logic shows why $x^2 =$ a negative has no real solution: nothing real squares to a negative, so there is no input to recover.

How the SOL examines this topic

• Fill-in-the-blank. Solve a square-root-type quadratic and type both solutions, exact radicals included.
• Multiple choice. Pick the solution set, with a distractor that keeps only the positive root.
• Drag-and-drop. Order the isolate-then-square-root steps.

Try this

Q1. Solve $x^2 = 81$. [1 point]

• Cue. $x = \pm 9$.

Q2. Solve $(x + 2)^2 = 25$. [2 points]

• Cue. $x + 2 = \pm 5$, so $x = 3$ or $x = -7$.