Solving Radical Equations

Learning goals

Isolate, raise, repeat, solve, then check every candidate
Trust that raising to a power loses no genuine solution
Explain extraneous roots as solutions of the sign-flipped twin or a widened domain
Read $\sqrt{t^2}$ as $\lvert t \rvert$ , so a square root is never negative
Recognize why the domain check alone cannot confirm a solution here
Note that cubing is reversible, unlike squaring

What raising both sides to a power does to the solution set

A radical equation is an equation in which the unknown sits under a radical, such as $\sqrt{x + 7} = x - 5$ . The unknown is trapped: you cannot add, subtract, or divide it out from under the root sign. The one operation that frees it is the operation that undoes the root, which is raising to a power.

That operation is legitimate, and here is exactly why. If two expressions are equal at some value of $x$ , they are the same number at that value. And the same number raised to the $n$ th power gives the same result. So for every positive integer $n$ ,

a = b \quad\Longrightarrow\quad a^n = b^n.

Read the arrow carefully, because everything in this lesson hangs on it. The implication runs one way. It does not run backwards when $n$ is even, and a single example settles that: $2^2 = 4$ and $(-2)^2 = 4$ , so $a^2 = b^2$ while $a \ne b$ .

The consequence for solving is precise. Take any value of $x$ that solves the original equation. Both sides are then equal numbers, so their $n$ th powers are equal too, so that same $x$ also solves the powered equation. Nothing that was a solution can be lost. What the arrow does not promise is the reverse, and so the powered equation may pick up values that the original never had. The solution set can grow but it can never shrink.

That is a useful bargain rather than a disaster. It means the candidates you find are a complete list: every genuine solution is somewhere among them. The list is just possibly too long. A value that solves the powered equation but not the original is called an extraneous root (or an extraneous solution). Your job is to find the candidates and then throw out the impostors.

So the method has five steps, and the last one is not a formality:

Isolate a radical, so that it stands alone on one side.
Raise both sides to the power that matches the index (square a square root, cube a cube root).
If a radical is still present, isolate and raise again.
Solve the radical-free equation.
Substitute every candidate into the original equation and keep only the ones that make it true.

Why squaring invents solutions

It is worth knowing exactly what the new equation says, not just that it says “something extra.” The factoring you learned for a difference of squares answers this completely.

Squaring both sides turns $A = B$ into $A = B$ or $A = -B$ #

Let $A$ and $B$ be real numbers. We claim that $A^2 = B^2$ holds if and only if $A = B$ or $A = -B$ .

Suppose first that $A^2 = B^2$ . Moving everything to one side gives $A^2 - B^2 = 0$ , and a difference of squares factors, so

(A - B)(A + B) = 0.

A product of two real numbers is zero only when one of the factors is zero, so either $A - B = 0$ or $A + B = 0$ . That is, either $A = B$ or $A = -B$ .

Now suppose the other direction. If $A = B$ , then squaring both sides gives $A^2 = B^2$ . If instead $A = -B$ , then $A^2 = (-B)^2 = B^2$ as well, since the square of a negative is the square of its opposite. Either way $A^2 = B^2$ .

Both directions hold, so the two statements say exactly the same thing about $A$ and $B$ .

Read that result as a statement about equations. When you square both sides of $A = B$ , you do not get a slightly different version of your equation. You get the equation

A = B \quad\text{or}\quad A = -B,

which is your equation glued together with its sign-flipped twin. Squaring cannot tell $t$ from $-t$ , so it cannot keep the two apart, and it hands back the solutions of both. When the two original sides were already real numbers, every extraneous root born this way is a solution of that twin. There is a second way an impostor can appear, and you will meet it shortly.

Now the puzzle from the opening dissolves. In $\sqrt{x} = -2$ we have $A = \sqrt{x}$ and $B = -2$ . Squaring produces ” $\sqrt{x} = -2$ or $\sqrt{x} = 2$ .” The first of those has no solutions, because the principal square root is never negative. The second gives $x = 4$ . So $x = 4$ did not come from the equation you asked about. It came from the twin that squaring folded in, free of charge.

The same fact has a compact form you should memorize:

\sqrt{t^2} = \lvert t \rvert, \qquad \text{which is } t \text{ only when } t \ge 0.

Squaring and then taking the principal root does not reliably return you to where you started. It returns the absolute value, because the principal root is non-negative by definition and cannot report a sign it was never given. When $t$ happens to be non-negative the two agree, and that is exactly what makes the identity treacherous. It looks like an inverse until the day $t$ is negative, and then it quietly hands back $-t$ instead. This is also why radical equations and absolute value equations misbehave in the same way. Taking principal roots on both sides of $A^2 = B^2$ gives $\lvert A \rvert = \lvert B \rvert$ , which is precisely $A = \pm B$ , the very split that created extraneous roots back in Chapter 1.

The solid curve is y = √x, the only thing the original equation talks about. Squaring adds the dashed lower branch y = -√x, and the line y = -2 meets that branch at x = 4. The candidate x = 4 is a perfectly ordinary number sitting comfortably inside the domain. It is simply on the branch that squaring invented.

Squaring has a second, quieter side effect. The powered equation is often defined at values where the original is meaningless. Look at $\sqrt{x - 5} = \sqrt{1 - x}$ . Squaring gives $x - 5 = 1 - x$ , a harmless linear equation with the single root $x = 3$ . But at $x = 3$ the original reads $\sqrt{-2} = \sqrt{-2}$ , and $\sqrt{-2}$ is not a real number, so $x = 3$ solves nothing. Squaring widened the domain, and a candidate wandered in through the new opening.

So a candidate can fail for two different reasons. It can be outside the domain of the original radicals, or it can be inside the domain and still make the two sides disagree in sign. Substituting into the original catches both at once, without your having to decide in advance which trap you fell into, because it simply asks whether the original statement is true.

Check your understanding

Squaring both sides of $\sqrt{x - 6} = \sqrt{3 - x}$ gives $x - 6 = 3 - x$ , so $x = 4.5$ . Why is $x = 4.5$ extraneous?

A Because

x = 4.5

is a solution of the sign-flipped twin

\sqrt{x - 6} = -\sqrt{3 - x}

, not of the original equation. B Because at

x = 4.5

both radicands are negative, so neither side of the original equation is a real number: squaring widened the domain and let this candidate in. C Because

x = 4.5

is not an integer, and only integer solutions are acceptable. D It is not extraneous;

x = 4.5

is a genuine solution.

Check your understanding

Squaring both sides of $\sqrt{x + 3} = x - 3$ gives $x + 3 = (x - 3)^2$ . Which statement about that step is correct?

A The squared equation might be missing a genuine solution of the original equation. B The squared equation is equivalent to "

\sqrt{x+3} = x - 3

\sqrt{x+3} = 3 - x

," the sign-flipped twin, so every genuine solution survives but an impostor might ride along. C Since

\sqrt{(x-3)^2} = x - 3

for every real

x

, squaring can always be undone directly, with no twin to worry about. D Because the radicand

x + 3

must be non-negative, every candidate that satisfies it automatically satisfies the original equation.

Isolate the radical before you raise it

Suppose you skip step 1 and square $\sqrt{3x + 1} - 2 = 3$ exactly as it stands. The left side is a difference of two terms, and squaring a difference produces a cross term:

\left(\sqrt{3x + 1} - 2\right)^2 = (3x + 1) - 4\sqrt{3x + 1} + 4.

The cross term $-4\sqrt{3x+1}$ still carries a radical, so you have done a lot of work and freed nothing. This is not a special misfortune. Squaring $P + Q$ always leaves the cross term $2PQ$ behind, and if $P$ is a radical then $2PQ$ contains that radical. The radical only disappears when it is alone, because then it is the whole side, and $(\sqrt{u})^2 = u$ with nothing left over.

Worked example 1 Solve $\sqrt{3x + 1} - 2 = 3$

Isolate the radical first by adding $2$ to both sides:

\sqrt{3x + 1} = 5.

Now the radical is the entire left side, so squaring clears it cleanly:

3x + 1 = 25.

Solve the linear equation: $3x = 24$ , so $x = 8$ .

Check the candidate in the original equation, not in any of the lines below it:

\sqrt{3(8) + 1} - 2 = \sqrt{25} - 2 = 5 - 2 = 3. \checkmark

The candidate survives, so $x = 8$ is the solution. Notice that the right side, $5$ , was non-negative when we squared, which is why no impostor appeared here.

The check is against the original, not against the domain

Five lessons ago, solving a rational equation, you also met extraneous roots, and there the check had a shortcut: throw out any candidate that makes a denominator zero. That shortcut is a domain check, and here it is not enough. A radical equation can hand you a candidate that is happily inside the domain, with every radicand non-negative, and it can still be an impostor.

Worked example 2 Solve $\sqrt{x + 7} = x - 5$

The radical already stands alone, so square both sides:

x + 7 = (x - 5)^2 = x^2 - 10x + 25.

Collect everything on one side to get a quadratic:

x^2 - 11x + 18 = 0.

This factors as $(x - 9)(x - 2) = 0$ , so the candidates are $x = 9$ and $x = 2$ .

Check $x = 9$ in the original. The left side is $\sqrt{9 + 7} = \sqrt{16} = 4$ , and the right side is $9 - 5 = 4$ . The two agree, so $x = 9$ is a genuine solution.

Check $x = 2$ in the original. The left side is $\sqrt{2 + 7} = \sqrt{9} = 3$ , and the right side is $2 - 5 = -3$ :

3 \ne -3.

So $x = 2$ is extraneous, and the only solution is $x = 9$ .

Look hard at the candidate that failed. At $x = 2$ the radicand is $2 + 7 = 9$ , so $x = 2$ sits comfortably inside the domain and both sides of the original are well-defined real numbers. It failed for one reason only: the left side came out $+3$ , the right side came out $-3$ . They agree in size and disagree in sign, exactly the disagreement squaring cannot see.

You can even see which equation $x = 2$ really belongs to. The twin of $\sqrt{x + 7} = x - 5$ is $\sqrt{x + 7} = -(x - 5) = 5 - x$ , and at $x = 2$ that reads $3 = 3$ . The candidate is an honest solution, just not of the equation you were asked about.

The moral is worth stating flatly, because it is the thing students most often get backwards. It is false that a candidate is extraneous exactly when it lies outside the domain. That rule belongs to rational equations, where the only irreversible move was multiplying by something that might be zero. Here the sign is what was thrown away, so the sign is what must be checked. And the only way to check it is to substitute into the original equation and look at both sides.

Check your understanding

Solve $\sqrt{x + 2} = x$ .

x = 2

x = 2

x = -1

x = -1

D no solution

Equations with two radicals

When two radicals appear, isolating one of them usually leaves a binomial like $1 + \sqrt{u}$ on the other side, and squaring that binomial gives $1 + 2\sqrt{u} + u$ : the $\sqrt{u}$ survives in the middle instead of disappearing. So you isolate that surviving radical and square a second time. One squaring is not enough whenever a radical ends up added to something else, which is the usual case, and it costs you two chances to invent a solution instead of one, so the final check matters even more here. (The one shortcut: if each side is nothing but a single isolated radical, as in $\sqrt{A} = \sqrt{B}$ , squaring clears both at once, since there is no sum to produce a cross term. It is the addition, not the count of radicals, that forces the second squaring.)

Worked example 3 Solve $\sqrt{2x + 3} - \sqrt{x + 1} = 1$

Isolate one radical by moving the other one across:

\sqrt{2x + 3} = 1 + \sqrt{x + 1}.

Square both sides. The right side is a binomial, so expand it with care:

2x + 3 = 1 + 2\sqrt{x + 1} + (x + 1) = x + 2 + 2\sqrt{x + 1}.

One radical is left, exactly as predicted. Isolate it by subtracting $x + 2$ from both sides:

x + 1 = 2\sqrt{x + 1}.

Square again:

(x + 1)^2 = 4(x + 1).

Bring everything to one side and factor out the common $x + 1$ :

(x + 1)\left[(x + 1) - 4\right] = 0 \;\Longrightarrow\; (x + 1)(x - 3) = 0.

The candidates are $x = -1$ and $x = 3$ .

Check $x = -1$ : the original becomes $\sqrt{1} - \sqrt{0} = 1 - 0 = 1$ , which is correct.

Check $x = 3$ : the original becomes $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$ , which is also correct.

Both candidates survive, so the solutions are $x = -1$ and $x = 3$ . Checking is not the same as rejecting: sometimes every candidate is honest, and a negative value of $x$ is perfectly acceptable as long as it keeps each radicand non-negative.

Check your understanding

Solve $\sqrt{x + 7} - \sqrt{x} = 1$ .

x = 4

x = 9

C no solution D

x = 1

Cubing both sides is reversible

Everything above blamed the even power. An odd power is a different animal, and the difference is not a technicality: it changes what you are allowed to conclude.

For real numbers, $a^3 = b^3$ holds exactly when $a = b$ #

One direction is immediate: if $a = b$ , then cubing both sides gives $a^3 = b^3$ .

For the other direction, suppose $a^3 = b^3$ . Then $a^3 - b^3 = 0$ , and a difference of cubes factors. (Expand the right side below and watch the middle terms cancel in pairs if you want to see why.)

(a - b)\left(a^2 + ab + b^2\right) = 0.

So either $a = b$ , which is what we want, or the second factor vanishes. Complete the square on that second factor to see when it can:

a^2 + ab + b^2 = \left(a + \tfrac{b}{2}\right)^2 + \tfrac{3}{4}b^2.

Both terms on the right are squares of real numbers, so both are non-negative, and their sum is zero only if each is zero separately. That forces $b = 0$ from the second term, and then $a = 0$ from the first. So the second factor vanishes only at $a = b = 0$ , where $a = b$ holds anyway.

In every case $a = b$ , so cubing loses nothing and adds nothing.

So cubing both sides of $A = B$ is reversible: the cubed equation says exactly what the original said, no more and no less. When a cube root stands alone, that one step is the whole method, and it cannot manufacture an impostor. Two familiar worries evaporate with it. A cube root is defined for every real number, positive or negative, so there is no domain to restrict. And $\sqrt[3]{u} = -3$ is a perfectly reasonable statement, because an odd root is allowed to be negative, unlike the principal square root. The same reasoning applies to fifth roots, seventh roots, and every other odd index. Every even index, by contrast, behaves just like the square root, because $t^4 = (-t)^4$ for the same reason $t^2 = (-t)^2$ .

Keep the guarantee inside its fence: it is a claim about the step “cube both sides of $A = B$ ”, not an amnesty for every equation that happens to contain a cube root. A solution that manipulates a cube-root equation some other way is not making that reversible move, and can still enlarge the solution set. Substituting back costs one line, so keep doing it.

Worked example 4 Solve $\sqrt[3]{2x - 1} = -3$

Do not reach for the “a root cannot be negative” reflex. That rule is about the principal square root. A cube root reports the sign of its input, so a negative value on the right is exactly what you should expect from a negative radicand.

Cube both sides, which is a reversible step:

2x - 1 = (-3)^3 = -27.

Solve: $2x = -26$ , so $x = -13$ .

Check in the original, which is still worth doing to catch arithmetic slips:

\sqrt[3]{2(-13) - 1} = \sqrt[3]{-27} = -3. \checkmark

The solution is $x = -13$ . Here the check confirmed rather than filtered, and by the proof above it could never have done anything else.

Rational exponents wear the same equation

You already know that $a^{1/n}$ is another name for $\sqrt[n]{a}$ , so an equation written with fractional exponents is a radical equation in different clothes. In particular $(2x + 5)^{1/2}$ is $\sqrt{2x + 5}$ , raising both sides to the power $2$ is squaring, and every warning in this lesson applies without a single change.

Worked example 5 Solve $(2x + 5)^{1/2} = x + 1$

Rewrite the left side to see what you are holding: $(2x + 5)^{1/2} = \sqrt{2x + 5}$ . The radical is already isolated, so raise both sides to the power $2$ , using the power-of-a-power rule on the left:

\left[(2x + 5)^{1/2}\right]^2 = (2x + 5)^{1} = 2x + 5.

The equation becomes

2x + 5 = (x + 1)^2 = x^2 + 2x + 1,

so $x^2 = 4$ and the candidates are $x = 2$ and $x = -2$ .

Check $x = 2$ : the left side is $(9)^{1/2} = 3$ and the right side is $2 + 1 = 3$ , so it is a solution.

Check $x = -2$ : the left side is $(1)^{1/2} = 1$ and the right side is $-2 + 1 = -1$ :

1 \ne -1.

So $x = -2$ is extraneous, and the only solution is $x = 2$ . Once again the impostor was inside the domain (its radicand was $1$ ) and failed only on sign.

When the numerator of the exponent is not $1$ , split the power into two steps rather than guessing. To solve $(x - 4)^{3/2} = 27$ , write the left side as $\left[(x-4)^{1/2}\right]^3$ and name the inner piece $u = (x - 4)^{1/2}$ . Because $u$ is a principal square root, $u \ge 0$ . The equation is now $u^3 = 27$ , and cubing is reversible, so $u = 3$ and nothing else. Then $(x - 4)^{1/2} = 3$ gives $x - 4 = 9$ , so $x = 13$ , and the check confirms it: $(13 - 4)^{3/2} = 9^{3/2} = (\sqrt{9})^3 = 27$ .

The same reading tells you when to stop early. The equation $(x - 4)^{3/2} = -27$ has no solution, because $u \ge 0$ forces $u^3 \ge 0$ , and $u^3$ can never equal $-27$ . Recognizing the impossibility before you compute saves the whole calculation.

Check your understanding

Solve $(x + 5)^{3/2} = 8$ .

x = -1

x = 59

x = 3

D no solution

Three causes, three checks

You have now met extraneous roots three times, and they had a different cause every time. That is the point worth carrying away. “Extraneous root” is not one phenomenon with one fix. It is a name for what happens whenever a step in your solution is not reversible. And since the irreversible step differs, the reason a false value slips in differs, and so the check that catches it differs.

Where you met it	The step that is not reversible	Why a false value slips in	What the check must ask
Absolute value equations	Rewriting $\lvert X \rvert = Y$ as $X = Y$ or $X = -Y$	The rewrite is only valid when $Y \ge 0$ , and that hypothesis is silently discarded	Does the value satisfy the original, signs and all?
Rational equations	Multiplying both sides by the LCD	The LCD can be zero, and multiplying by zero makes any two sides agree	Is the value in the domain, that is, does it leave every denominator non-zero?
Radical equations	Raising both sides to an even power	An even power cannot tell $t$ from $-t$ , so it folds in the twin equation $A = -B$	Does the value satisfy the original, signs and all?

A student who memorized “check for extraneous roots” without knowing why will reach for the wrong check. Coming from rational equations, the habit is to scan the candidates against the domain and keep whatever survives. Do that here and you will confidently report $x = 2$ as a solution of $\sqrt{x + 7} = x - 5$ , because its radicand is a healthy $9$ , and you will be wrong.

There is one check that never needs you to diagnose the cause first. Substitute the candidate into the original equation and see whether both sides come out equal. It works for absolute value equations, for rational equations, for radical equations, and for every irreversible step you will ever invent. That is because it is not a trick at all: it is the definition of what it means to be a solution.

Common mistakes

Squaring before isolating. Squaring $\sqrt{3x+1} - 2 = 3$ as it stands produces the cross term $-4\sqrt{3x+1}$ and frees nothing. Get the radical alone on one side first.
Skipping the check. The check is step 5 of the method, not an optional flourish. Without it you will report roots that solve nothing.
Checking against the domain instead of the original. That is the rational-equation rule, and it is false here. A candidate can have a perfectly non-negative radicand and still fail, because the two sides can disagree in sign.
Discarding negative values of $x$ on sight. A negative $x$ is fine whenever it keeps every radicand non-negative. In $\sqrt{2x + 3} - \sqrt{x + 1} = 1$ , the value $x = -1$ is a genuine solution. It is a negative side of the equation that causes trouble, not a negative unknown.
Writing $\sqrt{x^2} = x$ . The principal root is non-negative, so $\sqrt{x^2} = \lvert x \rvert$ .
Squaring term by term. $\left(\sqrt{a} + \sqrt{b}\right)^2$ is $a + 2\sqrt{ab} + b$ , not $a + b$ , and $(x + 1)^2$ is $x^2 + 2x + 1$ , not $x^2 + 1$ . Every squaring of a sum carries a cross term.
Stopping after one squaring. With two radicals you will usually still have one left. Isolate it and square again rather than declaring the job done.
Treating a cube root like a square root. $\sqrt[3]{u} = -3$ has an honest solution, because odd roots are allowed to be negative. It is $\sqrt{u} = -3$ that is impossible.

Key takeaways

To solve a radical equation: isolate a radical, raise both sides to the matching power, repeat if a radical remains, solve, then check every candidate in the original equation.
Raising both sides of an equation to a positive integer power never loses a genuine solution. When the power is even, that step is one-way: your candidate list is always complete, but it may be too long. When the power is odd, as cubing below shows, nothing extra gets added either.
Squaring turns $A = B$ into ” $A = B$ or $A = -B$ ”, because $A^2 = B^2$ factors as $(A - B)(A + B) = 0$ . Many extraneous roots are solutions of that sign-flipped twin; others appear because squaring widens the domain.
$\sqrt{t^2} = \lvert t \rvert$ , which equals $t$ only when $t \ge 0$ . The principal root is non-negative, which is why $\sqrt{\text{anything}}$ can never equal a negative number.
A candidate can sit safely inside the domain and still be extraneous. The domain check that worked for rational equations is not the check for radical equations.
Cubing both sides is reversible ( $a^3 = b^3$ exactly when $a = b$ ), so an isolated cube root raises none of squaring’s trouble, and a negative right-hand side is perfectly acceptable. The guarantee covers that step, not every equation containing a cube root, so keep substituting back.
Rational exponents are radicals in disguise: $(u)^{1/2}$ is $\sqrt{u}$ , and every warning above applies unchanged.
Substituting into the original equation is the one check that always works, because it is the definition of being a solution.

Solving Radical Equations

Learning goals

What raising both sides to a power does to the solution set

Why squaring invents solutions

Isolate the radical before you raise it

The check is against the original, not against the domain

Equations with two radicals

Cubing both sides is reversible

Rational exponents wear the same equation

Three causes, three checks

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice