Solving Radical Equations, SAT math lesson

Graph $y = \sqrt{x + 2}$ and the line $y = x$ , and you see them meet at exactly one point, $(2, 2)$ . Solve $\sqrt{x + 2} = x$ by squaring, though, and the algebra hands you two answers, $2$ and $-1$ . One of them is a ghost. Squaring is the right tool for removing a square root, but it answers a slightly different question than the one you asked, so every candidate has to go back into the original equation.

See also Solving Equations and Checking Solutions and The Square-Root Function in Foundations for Calculus and Solving Rational and Radical Equations in Precalculus.

Key terms

Radical equation

Definition A radical equation is an equation in which the variable appears inside a radical, such as a square root $\sqrt{\ }$ or a cube root $\sqrt[3]{\ }$ . A real number is a solution when every radical in the equation is defined as a real number at that value and the two sides are equal.

In plain words It is an equation with $x$ under a root sign. The usual plan is to get the root alone on one side, then raise both sides to a power that removes it.

Example Solve $\sqrt{x + 3} = 4$ . The root is already alone, so square both sides to get $x + 3 = 16$ , which gives $x = 13$ . Check $\sqrt{13 + 3} = \sqrt{16} = 4$ , which is correct.

Principal root

Definition For a real number $a \ge 0$ , the principal square root $\sqrt{a}$ is the unique nonnegative real number whose square is $a$ . So $\sqrt{a} \ge 0$ always, and for every real $x$

\sqrt{x^2} = \lvert x \rvert.

More generally, for an even index $n$ the symbol $\sqrt[n]{a}$ means the nonnegative $n$ th root of $a \ge 0$ . For an odd index the root is defined for every real $a$ and has the same sign as $a$ .

In plain words The square root sign always gives the nonnegative answer. The equation $x^2 = 25$ has two solutions, $5$ and $-5$ , but the number $\sqrt{25}$ is just $5$ .

Example The value $\sqrt{(-7)^2}$ is $\sqrt{49} = 7$ , not $-7$ . Odd roots behave differently. The equation $\sqrt[3]{x - 1} = -2$ is fine with a negative right side. Cubing gives $x - 1 = -8$ , so $x = -7$ , and indeed

\sqrt[3]{-8} = -2.

Domain restriction

Definition In the real numbers, an even root such as $\sqrt{E}$ is defined only when the radicand $E$ is nonnegative. In an equation of the form

\sqrt{E} = F,

two conditions follow. The radicand needs $E \ge 0$ , and since a principal square root is never negative, a solution also needs $F \ge 0$ . If $F$ is a negative constant, the equation has no real solution.

In plain words The inside of a square root must not be negative, and the root itself is never negative. So if the root equals something, that something must be zero or positive.

Example Solve $\sqrt{x - 4} = x - 6$ . A solution needs $x - 4 \ge 0$ and $x - 6 \ge 0$ , so $x \ge 6$ . Squaring gives

\begin{aligned} x - 4 &= x^2 - 12x + 36 \\[0.4em] 0 &= x^2 - 13x + 40 \\[0.4em] 0 &= (x - 5)(x - 8). \end{aligned}

The candidate $5$ fails $x \ge 6$ , since $\sqrt{1} = 1$ but $5 - 6 = -1$ . The candidate $8$ works, since $\sqrt{4} = 2 = 8 - 6$ . The only solution is $8$ .

As a second case, $\sqrt{2x + 1} + 5 = 2$ means $\sqrt{2x + 1} = -3$ , which no real $x$ can satisfy. The equation has no solution, and you can say so without squaring at all.

Squaring both sides

Definition For real numbers $a$ and $b$ , if $a = b$ then $a^2 = b^2$ . The converse is false, because

\begin{gathered} a^2 = b^2 \quad \text{means} \\[0.4em] a = b \quad \text{or} \quad a = -b. \end{gathered}

So squaring both sides of an equation keeps every solution, but the squared equation can also have solutions of $a = -b$ . The squared equation alone does not establish that a candidate solves the original. Squaring both sides of $\sqrt{E} = F$ is reversible exactly when $F \ge 0$ .

In plain words Squaring loses the sign. Since $3 \ne -3$ but $3^2 = (-3)^2$ , an equation that was false can become true after squaring.

Example Solve $\sqrt{x + 10} + 2 = x$ . First isolate the root, $\sqrt{x + 10} = x - 2$ . Then square and solve,

\begin{aligned} x + 10 &= x^2 - 4x + 4 \\[0.4em] 0 &= x^2 - 5x - 6 \\[0.4em] 0 &= (x - 6)(x + 1). \end{aligned}

The candidates are $6$ and $-1$ . Squaring before isolating would have left a messy cross term with the root still inside, which is why isolating comes first.

Extraneous solution

Definition An extraneous solution of a radical equation is a candidate from the squared (or otherwise raised) equation that does not satisfy the original equation. It appears when the candidate makes the two sides opposites, $a = -b$ with $a \ne b$ , or when it makes a radicand negative. The only reliable test is to substitute each candidate into the original equation.

In plain words It is an answer that the squared equation accepts but the original equation rejects. Plug in, compute both sides, and keep only the candidates where they truly match.

Example Return to $\sqrt{x + 10} + 2 = x$ with candidates $6$ and $-1$ .

Candidate	Left side	Right side	Result
$6$	$\sqrt{16} + 2 = 6$	$6$	solution
$-1$	$\sqrt{9} + 2 = 5$	$-1$	extraneous

The only solution is $6$ . In the opening equation $\sqrt{x + 2} = x$ the same thing happens. Squaring gives $x + 2 = x^2$ , so $(x - 2)(x + 1) = 0$ , but at $x = -1$ the left side is $\sqrt{1} = 1$ while the right side is $-1$ . Only $2$ is a solution, which matches the single intersection of the two graphs. In the Desmos graphing calculator you can graph $y = \sqrt{x + 2}$ and $y = x$ and click the intersection point to read its coordinates.

Finding an input from a radical rule

Definition For a function such as $f(x) = a\sqrt{x - h} + k$ with $a \ne 0$ , the inputs with $f(x) = c$ are the solutions of that radical equation in the domain $x \ge h$ . Isolate the root, use the sign condition on the isolated side, square, and check.

In plain words When a question gives the output of a square root rule and asks for the input, undo the outside steps first, then square.

Example Let $f(x) = 3\sqrt{x - 2}$ . To find $x$ with $f(x) = 12$ , divide by $3$ to get $\sqrt{x - 2} = 4$ . Squaring gives $x - 2 = 16$ , so $x = 18$ . Check $3\sqrt{16} = 12$ . For $f(x) = -6$ the isolated root would equal $-2$ , so no input gives that output.

Check yourself

What is the solution set of $\sqrt{x + 6} = x$ ? (A) $\{3\}$ (B) $\{-2\}$ (C) $\{-2, 3\}$ (D) There is no solution.

Show the answer

The answer is (A). Squaring gives $x + 6 = x^2$ , so $(x - 3)(x + 2) = 0$ and the candidates are $3$ and $-2$ . At $x = 3$ both sides equal $3$ . At $x = -2$ the left side is $\sqrt{4} = 2$ and the right side is $-2$ , so $-2$ is extraneous. Choice (C) skips the check, choice (B) keeps the wrong candidate, and choice (D) rejects a valid one.

If $\sqrt{3x - 2} + 4 = 9$ , what is the value of $x$ ?

Show the answer

Subtract $4$ to isolate the root, $\sqrt{3x - 2} = 5$ . The right side is positive, so squaring is safe, and $3x - 2 = 25$ gives $x = 9$ . Check $\sqrt{25} + 4 = 9$ .

Which of the following equations has no real solution? (A) $\sqrt{x - 5} = 0$ (B) $\sqrt{x + 1} + 3 = 1$ (C) $\sqrt{x} = x$ (D) $\sqrt{4 - x} = 2$

Show the answer

The answer is (B). Isolating the root gives $\sqrt{x + 1} = -2$ , and a principal square root is never negative. Squaring would suggest $x = 3$ , but $\sqrt{4} + 3 = 5$ , not $1$ . The others all have solutions. Choice (A) is solved by $x = 5$ , choice (C) by $0$ and $1$ , and choice (D) by $0$ .

Solving Radical Equations

Theory, part 63

Practice, part 64

Key terms

Radical equation

Principal root

Domain restriction

Squaring both sides

Extraneous solution

Finding an input from a radical rule

Check yourself

How this lesson connects

Builds on