Simplifying Radical Expressions

Intuition

You have been writing $\sqrt{\phantom{x}}$ for chapters now: in the quadratic formula, in the discriminant, in the modulus of a complex number, in the irrational roots of a polynomial. The last lesson finally pinned down what the symbol means, in order to define $a^{1/n}$ , but that definition was moving fast toward a different destination. This lesson slows down and puts it to work. One of the rules you have been using on trust, $\sqrt{ab} = \sqrt a \sqrt b$ , is false as soon as $a$ and $b$ are both negative, and you now own enough machinery to watch it fail. Every result here, including the identity $\sqrt{x^2} = |x|$ that trips up more students than any other in algebra, follows from the same decision the last lesson made: the radical sign must name exactly one number.

Learning goals

Read $\sqrt a$ as the principal root, one number not two
Write $\sqrt[n]{x^n}$ as $\lvert x \rvert$ for even $n$
Apply the product and quotient rules, nonnegative for even index, nonzero divisor for odd
Identify the three conditions for simplest radical form
Reduce a common factor of index and exponent
Reject $\sqrt{a+b} = \sqrt a + \sqrt b$

One symbol, one number

Start with a question you can already answer. The equation

x^2 = 9

has two real solutions, $x = 3$ and $x = -3$ . Both of them square to $9$ , so both of them deserve the name “a square root of $9$ .” That is a fact about the equation, and nothing in this lesson changes it.

Now ask a different question: what number does the symbol $\sqrt 9$ stand for? It cannot stand for both of them. A symbol that might mean $3$ or might mean $-3$ is useless, because then $\sqrt 9 + \sqrt 9$ could be $6$ , or $0$ , or $-6$ . With that ambiguity in play, no equation containing a radical would say anything definite. A symbol has to name one definite value to be useful at all.

So a choice was made, and it is the foundation of everything below. It is the same choice the last lesson made in order to define $a^{1/n}$ ; here it is restated on its own, because everything from this point on depends on it directly.

So $\sqrt 9 = 3$ , and only $3$ . The other square root of $9$ has to be written out as $-\sqrt 9 = -3$ . Two consequences follow at once, and they are worth saying out loud:

A real square root is never negative. Whenever $\sqrt{\text{stuff}}$ is a real number, it equals something $\ge 0$ . This one sentence will decide the whole next section.
“Solve $x^2 = 9$ ” and “evaluate $\sqrt 9$ ” are different questions. The first has two answers, the second has one.

That difference is exactly why the quadratic formula carries a $\pm$ in front of its radical:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

The radical hands back only the non-negative root, so the formula has to ask for the other one explicitly. The $\pm$ you have been writing since chapter 4 is not decoration. It is there to repair, by hand, the second root that the symbol deliberately throws away.

The word “unique” in the definition is doing real work, so here is why it holds. Picture $r$ starting at $0$ and growing. As $r$ grows, $r^n$ grows too, sweeping upward through every non-negative value exactly once on the way, never doubling back. So $r^n$ can equal $a$ at only one value of $r$ : there is exactly one non-negative number whose $n$ -th power is $a$ , which is what makes calling it the principal root legitimate.

Check your understanding

The equation $x^2 = 49$ has two real solutions. What is the value of the expression $\sqrt{49}$ ?

7

only B

-7

only C

7

and

-7

\pm 7

Roots of any index

Everything above generalizes. In the expression $\sqrt[n]{a}$ , the number $n$ is the index and $a$ is the radicand. A square root is the case $n = 2$ , where the index is left unwritten.

The definition splits into two cases, and the split is caused entirely by signs.

Odd index. Cube every real number and you get every real number back: $2^3 = 8$ , $(-2)^3 = -8$ . An odd power preserves sign, so a negative number has a negative cube root and a positive number has a positive one. In each of those cases there is exactly one real root. No choice has to be made and nothing has to be excluded:

\sqrt[3]{8} = 2, \qquad \sqrt[3]{-8} = -2, \qquad \sqrt[5]{-32} = -2.

Even index. An even power destroys sign: $r^4 \ge 0$ for every real $r$ . So a negative radicand has no real even root at all, and a positive radicand has two of them, $\pm r$ . Both problems get the same treatment as before: negative radicands are excluded, and of the two roots the non-negative one is chosen. For an even index, $\sqrt[n]{a}$ names a real number exactly when $a \ge 0$ , and that number is never negative:

\sqrt[4]{16} = 2, \qquad \sqrt[6]{64} = 2, \qquad \sqrt[4]{-16} \ \text{is not a real number}.

The last one is not meaningless, and chapter 5 is the reason. You know that $\sqrt{-1} = i$ , so a negative radicand under a square root has a perfectly good complex value. Keep the distinction sharp: for the rest of this lesson, every rule involving an even index is stated over the real numbers with a non-negative radicand. An odd index carries no such restriction, as the cube roots above already showed. Hold onto that, because the next section is where ignoring it, for an even index, costs you.

The square root of a square is an absolute value

Here is the single most important identity in the lesson, and the one most often written down wrong. Simplify $\sqrt{x^2}$ .

The tempting answer is $x$ : the square and the root should undo each other. Test it at $x = -3$ :

\sqrt{(-3)^2} = \sqrt{9} = 3, \qquad \text{but} \qquad x = -3.

So $\sqrt{x^2} = x$ is simply false at $x = -3$ . And you should have seen it coming, because of the first consequence of the definition: $\sqrt{\phantom{x}}$ never returns a negative number, while $x$ is negative half the time. An expression that is always $\ge 0$ cannot equal an expression that is sometimes $< 0$ .

What you want is the thing that leaves non-negative numbers alone and flips negative ones. You have had a name for it since chapter 1.

For every real $x$ , $\sqrt{x^2} = |x|$ #

By definition, $\sqrt{x^2}$ is the non-negative real number whose square is $x^2$ , and there is only one such number, by the uniqueness fact established above. So the whole job is to show that $|x|$ has those two properties, and uniqueness then forces $|x|$ to be that number.

First, $|x| \ge 0$ . That is immediate from the definition of absolute value, which reports a distance from zero.

Second, $|x|^2 = x^2$ . Check the two cases the absolute value splits into. If $x \ge 0$ , then $|x| = x$ , so $|x|^2 = x^2$ . If $x < 0$ , then $|x| = -x$ , so

|x|^2 = (-x)^2 = (-1)^2 x^2 = x^2.

Either way $|x|^2 = x^2$ .

So $|x|$ is a non-negative number whose square is $x^2$ . The principal square root of $x^2$ is the only non-negative number whose square is $x^2$ . Therefore $\sqrt{x^2} = |x|$ , for every real $x$ , with no exceptions and no side conditions.

Notice what this does and does not say. It does not say the square and the root fail to undo each other; it says they undo each other up to sign. The absolute value is the bookkeeping that fixes the sign. The two agree precisely on the non-negative numbers: $\sqrt{x^2} = x$ holds exactly when $x \ge 0$ . (In one direction, $x \ge 0$ gives $|x| = x$ . In the other, $\sqrt{x^2} = x$ makes $x$ equal to something non-negative, so $x \ge 0$ . Both directions check out, which is what “exactly when” demands.)

The graph of y = sqrt(x squared) is the V, and the graph of y = x is the straight line. They lie on top of each other for x at least 0, and separate for negative x: at x = -3 the radical returns 3, while x itself is -3. The V is the graph of the absolute value.

The same reasoning runs at every index, and the index’s parity decides the answer:

\sqrt[n]{x^n} = |x| \ \ (n \text{ even}), \qquad \sqrt[n]{x^n} = x \ \ (n \text{ odd}).

An even root is forced to be non-negative, so it needs the bars. An odd root is free to be negative and carries the sign of $x$ by itself, so bars would be wrong: $\sqrt[3]{(-2)^3} = \sqrt[3]{-8} = -2$ , exactly $x$ , not $|x|$ .

One refinement saves a lot of unnecessary bars. When you pull an even power apart, what matters is the exponent that comes out:

\sqrt{x^4} = x^2, \qquad \sqrt{x^6} = |x|^3, \qquad \sqrt{x^{10}} = |x|^5 .

In the first, $x^2$ is already non-negative for every real $x$ , so bars would change nothing and are left off. In the others the exponent coming out is odd, so the result could be negative and the bars are mandatory. Bars are needed when, and only when, the exponent that emerges is odd (and the variable is unrestricted).

Worked example 1 Simplify $\sqrt{20x^2}$ , where $x$ is any real number

Check the domain first. The radicand is $20x^2$ , and $x^2 \ge 0$ for every real $x$ , so $20x^2 \ge 0$ always. Nothing restricts $x$ , which means the answer has to be correct for negative $x$ too.

Split off the largest perfect square inside, which is $4x^2$ :

\sqrt{20x^2} = \sqrt{4x^2 \cdot 5} = \sqrt{4x^2}\,\sqrt{5}.

(The product rule is doing that split, and both factors are non-negative, so it is allowed. The next section is about what happens when they are not.) Now $\sqrt{4x^2} = \sqrt{2^2}\sqrt{x^2} = 2|x|$ , using the identity just proved:

\sqrt{20x^2} = 2|x|\sqrt 5 .

The bars are not optional here. Test $x = -1$ : the original is $\sqrt{20} \approx 4.47$ , a positive number. The candidate answer $2x\sqrt 5$ would give $-2\sqrt5 \approx -4.47$ , the wrong sign. The correct answer $2|x|\sqrt5$ gives $2\sqrt5 \approx 4.47$ , which matches.

Check your understanding

Simplify $\sqrt{9x^2}$ , where $x$ may be any real number.

3x

3|x|

9|x|

\pm 3x

The product rule, and the hypothesis it depends on

The rule you have been using on trust is the product rule, and it comes with a hypothesis attached:

\sqrt{ab} = \sqrt a \, \sqrt b, \qquad \text{provided } a \ge 0 \text{ and } b \ge 0 .

That hypothesis is usually skated over, as though it were a formality. Watch where it enters the proof, because in a moment you are going to remove it and watch the rule die.

If $a, b \ge 0$ , then $\sqrt{ab} = \sqrt a \sqrt b$ #

Write $p = \sqrt a$ and $q = \sqrt b$ . These exist as real numbers because $a$ and $b$ are non-negative, which is the first use of the hypothesis.

By the definition of the principal root, $p \ge 0$ and $q \ge 0$ , and also $p^2 = a$ and $q^2 = b$ . Their product $pq$ then satisfies two things. Its square is

(pq)^2 = p^2 q^2 = ab,

and it is non-negative, since $p \ge 0$ and $q \ge 0$ make $pq \ge 0$ . That is the second use of the hypothesis, and it is the one that matters.

So $pq$ is a non-negative real number whose square is $ab$ . By the uniqueness fact established earlier, there is only one such number, and its name is $\sqrt{ab}$ . Therefore $pq = \sqrt{ab}$ , which is exactly $\sqrt a \sqrt b = \sqrt{ab}$ .

The proof splits into two claims: that $pq$ squares to $ab$ , and that $pq$ is non-negative. The first claim never uses the sign of $a$ or $b$ at all. The second one does, and it is the claim that identifies $pq$ as the principal root rather than merely a square root. Pull the hypothesis and the first half of the argument survives while the second half collapses. So the product $pq$ lands on a square root of $ab$ , just possibly the wrong one of the two.

That is a prediction, and you have the tools to test it.

Watching the rule fail

Take $a = b = -1$ . Compute both sides.

The left side is easy, and it is entirely real:

\sqrt{ab} = \sqrt{(-1)(-1)} = \sqrt{1} = 1 .

The right side uses the imaginary unit from chapter 5, where $\sqrt{-1} = i$ and $i^2 = -1$ :

\sqrt{a}\,\sqrt{b} = \sqrt{-1}\cdot\sqrt{-1} = i \cdot i = i^2 = -1 .

One side is $1$ and the other is $-1$ . The product rule is not “risky” here, or “undefined,” or a matter of taste. For $a = b = -1$ it is false, and you have just checked it with arithmetic you already own.

The diagnosis is exactly what the proof predicted. The number $-1$ that the right side produced is a square root of $ab = 1$ , because $(-1)^2 = 1$ . It is simply the wrong square root: it is the negative one, and $\sqrt{\phantom{x}}$ is committed to the non-negative one. The hypothesis $a, b \ge 0$ is what guarantees the product $pq$ lands on the correct root, and with two negative factors that guarantee is gone.

The failure is not a one-off:

\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6, \qquad \text{while} \qquad \sqrt{(-4)(-9)} = \sqrt{36} = 6 .

Off by a sign again. With one negative factor there is no problem at all: $\sqrt{-4}\cdot\sqrt{9} = (2i)(3) = 6i$ , and $\sqrt{(-4)(9)} = \sqrt{-36} = 6i$ agrees. It takes two negative factors to break the rule, and two negative factors break it every time, always by a factor of $-1$ .

The practical rule that follows is short, and it is the one to carry into every problem with a negative radicand: convert to $i$ first, then multiply. Never combine two negative radicands under one radical sign.

Check your understanding

What is $\sqrt{-4}\cdot\sqrt{-25}$ ?

10

-10

10i

-10i

Nothing in that proof cared that the index was $2$ . It needed only that a non-negative radicand has exactly one non-negative $n$ -th root, which holds for every $n$ , as established earlier. Replace each square root by an $n$ -th root and the argument runs word for word, so the rule holds at every index:

\sqrt[n]{ab} = \sqrt[n]{a}\,\sqrt[n]{b}, \qquad \text{provided } a \ge 0 \text{ and } b \ge 0 .

That is the step you take whenever you pull a fourth root or a sixth root apart, as in $\sqrt[4]{16x^8} = \sqrt[4]{16}\,\sqrt[4]{x^8}$ .

For an odd index none of the trouble with signs arises in the first place: $\sqrt[3]{a}$ is defined for every real $a$ and carries the sign of $a$ . So the odd-index product rule $\sqrt[3]{ab} = \sqrt[3]{a}\,\sqrt[3]{b}$ holds for all real $a$ and $b$ , negatives included. There is no second root competing for the name, so there is nothing to get wrong.

The quotient rule

Division behaves the same way, and the proof is the same argument with a quotient in place of a product. If $a \ge 0$ and $b > 0$ , then $\sqrt a / \sqrt b$ is non-negative and its square is $a/b$ , so it is the principal square root of $a/b$ :

\sqrt{\frac{a}{b}} = \frac{\sqrt a}{\sqrt b}, \qquad a \ge 0, \ b > 0 .

The condition $b > 0$ is stronger than $b \ge 0$ only because you cannot divide by zero. For an even index, the same statement holds with a non-negative $a$ and a positive $b$ :

\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}, \qquad n \text{ even}, \ a \ge 0, \ b > 0 .

For an odd index there is no nonnegative-output convention to protect, exactly as with the product rule, so the restriction relaxes to the one thing you can never do, dividing by zero: $a$ can be any real number and $b$ only has to be nonzero.

\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}, \qquad n \text{ odd}, \ b \ne 0 .

Both rules are worth reading in both directions. Left to right, they take a radical apart: $\sqrt{9/25} = 3/5$ . Right to left, they put one together, which is often the faster move:

\frac{\sqrt{50}}{\sqrt{2}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5 .

Check your understanding

What is $\sqrt{48} / \sqrt{3}$ ?

4

16

\sqrt{51}

144

Simplest radical form

Radicals, like fractions, have a standard form, so that two people who simplify the same expression write down the same thing. A radical expression is in simplest radical form when all three of these hold.

No perfect $n$ -th power factors remain under the radical. $\sqrt{72}$ fails this, because $36$ divides $72$ .
No radical appears in a denominator. $\dfrac{1}{\sqrt 2}$ fails this.
The index and the exponents under the radical share no common factor. $\sqrt[6]{x^4}$ fails this, because $6$ and $4$ share the factor $2$ .

Condition 2 is a real condition, but its repair is a technique of its own, and it closes this chapter (Rationalizing and Radical Conjugates). Leave it alone for now; every exercise here is arranged so that conditions 1 and 3 are the whole job.

Check your understanding

Which of these is already in simplest radical form?

\sqrt{18}

\sqrt[6]{8}

\dfrac{5}{\sqrt3}

2\sqrt5

Condition 1: extract perfect powers. Factor the radicand, pull out everything that appears as a perfect $n$ -th power, and leave the rest inside. Two routes work. Hunt for the largest perfect $n$ -th power that divides the radicand, which is fast when you spot it. The other route is to factor into primes and take out one factor for every group of $n$ equal ones, which never fails.

Worked example 2 Put $\sqrt{72}$ in simplest radical form

Route 1: the largest perfect square. The perfect squares are $1, 4, 9, 16, 25, 36, 49, \ldots$ , and the largest one dividing $72$ is $36$ , since $72 = 36 \cdot 2$ . Both factors are non-negative, so the product rule applies:

\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36}\,\sqrt{2} = 6\sqrt 2 .

Route 2: prime factorization. If the largest square is not obvious, break the radicand into primes and group them in pairs, since a pair of equal primes is a perfect square:

72 = 2^3 \cdot 3^2 = (2 \cdot 2)(3 \cdot 3)(2).

Each pair sends one copy of itself outside the radical, and the leftover $2$ stays in:

\sqrt{72} = 2 \cdot 3 \cdot \sqrt 2 = 6\sqrt 2 .

Both routes agree, as they must. A quick sanity check: $6\sqrt2 \approx 6(1.414) = 8.49$ , and $8.49^2 \approx 72$ .

The answer $6\sqrt2$ is not “less simplified” than $\sqrt{72}$ because it looks longer. It is more useful. It separates the exact whole-number part from the irrational part, which exposes that $\sqrt{72}$ and $\sqrt{2}$ are multiples of the same irrational number. Two expressions written in simplest form can be compared at a glance, and the next lesson turns that into arithmetic.

At a higher index, the grouping size changes but nothing else does. For a cube root you take out one factor per group of three.

Worked example 3 Simplify $\sqrt[3]{-54a^4}$ , where $a$ is any real number

The index is odd, so a negative radicand is fine and no absolute values will appear. Factor the radicand, splitting off perfect cubes:

-54a^4 = (-27)(2)\,(a^3)(a),

since $-54 = -27 \cdot 2$ and $a^4 = a^3 \cdot a$ . Both $-27 = (-3)^3$ and $a^3$ are perfect cubes, so they come out:

\sqrt[3]{-54a^4} = \sqrt[3]{-27}\,\sqrt[3]{a^3}\,\sqrt[3]{2a} = -3 \cdot a \cdot \sqrt[3]{2a} = -3a\sqrt[3]{2a} .

No bars anywhere. $\sqrt[3]{a^3} = a$ exactly, for every real $a$ , because an odd root keeps the sign it was given. Writing $|a|$ here would be an error, not a precaution: at $a = -1$ the original is $\sqrt[3]{-54} \approx -3.78$ , and $-3a\sqrt[3]{2a} = 3\sqrt[3]{-2} \approx -3.78$ matches, while $-3|a|\sqrt[3]{2a}$ would give $+3.78$ .

Worked example 4 Simplify $\sqrt{50x^3y^4}$ , where $x \ge 0$

Start with the domain, because it does half the work. The index is even, so the radicand must be non-negative. With $x \ge 0$ , every factor here is already non-negative for any real $y$ : $50 > 0$ , $x^3 \ge 0$ , and $y^4 \ge 0$ . So the expression is a real number no matter what $y$ is, and that is the hypothesis the rest of this example leans on.

Split each factor into a perfect square times a remainder:

50x^3y^4 = (25 \cdot 2)(x^2 \cdot x)(y^4),

so the perfect squares are $25$ , $x^2$ , and $y^4$ , and what stays behind is $2x$ :

\sqrt{50x^3y^4} = \sqrt{25}\,\sqrt{x^2}\,\sqrt{y^4}\,\sqrt{2x} = 5 \cdot |x| \cdot y^2 \cdot \sqrt{2x} .

Now clean up the two absolute values that could have appeared. Since $x \ge 0$ , $|x| = x$ . And $\sqrt{y^4} = y^2$ needs no bars in the first place, because the exponent coming out is even and $y^2 \ge 0$ for every real $y$ :

\sqrt{50x^3y^4} = 5xy^2\sqrt{2x} .

The general habit is worth naming. Ask what the domain forces before you ask whether you need bars. Bars are needed only for a variable whose sign is genuinely unknown, and an odd power under an even root usually pins the sign down for you.

Condition 3: reduce the index. Rational exponents, from the previous lesson, make this mechanical. Rewrite the radical as a power, reduce the fraction, and translate back.

Worked example 5 Simplify $\sqrt[6]{8}$ and $\sqrt[4]{9}$

For $\sqrt[6]{8}$ , write the radicand as a power of $2$ and convert to a rational exponent:

\sqrt[6]{8} = \sqrt[6]{2^3} = \left(2^3\right)^{1/6} = 2^{3/6} = 2^{1/2} = \sqrt 2 .

The index and the exponent shared the factor $3$ , and reducing $3/6$ to $1/2$ is precisely what canceling it means. A sixth root of $8$ really is $\sqrt2$ : check by raising it to the sixth power, $(\sqrt2)^6 = (2^{1/2})^6 = 2^3 = 8$ .

The same move handles $\sqrt[4]{9}$ :

\sqrt[4]{9} = \sqrt[4]{3^2} = 3^{2/4} = 3^{1/2} = \sqrt 3 .

For a variable base, keep the base non-negative so the rational-exponent rules apply. With $x \ge 0$ ,

\sqrt[6]{x^4} = x^{4/6} = x^{2/3} = \sqrt[3]{x^2},

which is in simplest form, since $3$ and $2$ share no common factor.

Check your understanding

Simplify $\sqrt[8]{16}$ .

\sqrt 2

2

4

16

One rule that does not exist

The product and quotient rules distribute a radical over multiplication and division, under the sign conditions above. It is tempting to assume the same courtesy for addition, but no such rule exists. $\sqrt{a + b}$ is not $\sqrt a + \sqrt b$ , for any $a$ and $b$ where both sides are defined and unequal to the trivial case. Test it once and the gap is obvious:

\sqrt{9 + 16} = \sqrt{25} = 5, \qquad \text{while} \qquad \sqrt 9 + \sqrt{16} = 3 + 4 = 7.

Five is not seven. There is no algebra to fix, because there was never a rule here to apply; addition and subtraction simply do not distribute over a radical the way multiplication and division do.

Check your understanding

What is $\sqrt{16 + 9}$ ?

5

7

\sqrt7

25

Common mistakes

Writing $\sqrt{x^2} = x$ . This is the big one. A square root is never negative, so it cannot equal $x$ when $x$ is negative. The identity is $\sqrt{x^2} = |x|$ , and it costs one pair of bars to be right: at $x = -3$ , $\sqrt{(-3)^2} = 3$ , not $-3$ .
Using bars where they do not belong. For an odd index there is no sign to fix: $\sqrt[3]{x^3} = x$ , bars and all would be wrong. And if the exponent that comes out is even, as in $\sqrt{x^4} = x^2$ , the result is already non-negative and the bars are redundant. Bars go on exactly when an odd exponent emerges from an even root taken over an unrestricted variable.
Applying $\sqrt{ab} = \sqrt a \sqrt b$ with two negative factors. $\sqrt{-4}\cdot\sqrt{-9}$ is $-6$ , not $6$ . Convert each radical to $i$ form before multiplying. This rule has a hypothesis, and the hypothesis is load-bearing.
Saying $x^2 = 9$ has one solution, or that $\sqrt 9 = \pm 3$ . Keep the two questions apart. The equation has two solutions ( $3$ and $-3$ ); the symbol $\sqrt 9$ names one number ( $3$ ). The $\pm$ in the quadratic formula exists to put the second one back.
Splitting a sum. There is no rule $\sqrt{a + b} = \sqrt a + \sqrt b$ , and it is not close to true: $\sqrt{9 + 16} = \sqrt{25} = 5$ , while $\sqrt 9 + \sqrt{16} = 3 + 4 = 7$ . Radicals distribute over products and quotients under the conditions above; there is no such rule for sums or differences.
Stopping too early. $\sqrt{72} = 2\sqrt{18}$ is true but not simplified, because $18$ still hides the perfect square $9$ . Take out the largest perfect power, or factor into primes so nothing can hide.

Key takeaways

$\sqrt a$ denotes the principal root: the one non-negative number whose square is $a$ . For $a > 0$ , the equation $x^2 = a$ has two real solutions, but the symbol names only one of them; for $a = 0$ there is just the single solution $x = 0$ .
$\sqrt{x^2} = |x|$ for every real $x$ , and $\sqrt{x^2} = x$ exactly when $x \ge 0$ . In general $\sqrt[n]{x^n} = |x|$ for even $n$ and $\sqrt[n]{x^n} = x$ for odd $n$ .
An odd root accepts any real radicand and keeps its sign; an even root needs a non-negative radicand (over the reals) and returns a non-negative value.
Product and quotient rules: $\sqrt[n]{ab} = \sqrt[n]a\sqrt[n]b$ and $\sqrt[n]{a/b} = \sqrt[n]a / \sqrt[n]b$ need $a \ge 0$ and $b \ge 0$ (with $b > 0$ for the quotient) at an even index; at an odd index both hold for every real $a$ , with only $b \ne 0$ required for the quotient. The even-index hypothesis is load-bearing, not decorative: with both factors negative the product rule is false at a square root, since $\sqrt{-1}\sqrt{-1} = -1$ while $\sqrt{(-1)(-1)} = 1$ . Convert to $i$ first, then multiply.
Simplest radical form: no perfect $n$ -th power factors left inside, no radical in a denominator, and no common factor between the index and the exponents. Reduce that common factor with rational exponents, as in $\sqrt[6]{8} = 2^{3/6} = \sqrt 2$ .
There is no rule for sums. $\sqrt{a+b}$ is not $\sqrt a + \sqrt b$ .

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Rational exponents and radicals (Khan Academy)
Simplifying square roots (Math is Fun)
Simplify expressions with roots (free textbook) (OpenStax)
Radicals: nth roots, properties, and simplest form (Paul's Online Math Notes)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Why a non-negative number has exactly one non-negative n-th root

A non-negative number has exactly one non-negative $n$ -th root#

Fix an integer $n \ge 2$ and a real number $a \ge 0$ . That some non-negative real $r$ satisfies $r^n = a$ is a fact about the completeness of the real numbers. As $r$ climbs from $0$ upward, $r^n$ climbs continuously from $0$ upward and passes through every non-negative value on the way, so it passes through the non-negative number $a$ . What needs proving is that it passes through $a$ only once.

Suppose two non-negative reals $r$ and $s$ both satisfy $r^n = s^n = a$ , and suppose they are different. Relabel them if necessary so that $r < s$ . Now use a factorization you met with polynomials:

s^n - r^n = (s - r)\left(s^{n-1} + s^{n-2}r + \cdots + sr^{n-2} + r^{n-1}\right).

Look at the two factors on the right. The first, $s - r$ , is positive, because $r < s$ . The second is a sum of products of non-negative numbers, so no term in it is negative, and its leading term $s^{n-1}$ is strictly positive because $s > r \ge 0$ forces $s > 0$ . A sum of non-negative terms with at least one positive term is positive.

A positive number times a positive number is positive, so $s^n - r^n > 0$ , which says $s^n \ne r^n$ . That contradicts $r^n = s^n = a$ . No two distinct non-negative reals can share an $n$ -th power, so the non-negative $n$ -th root of $a$ is unique, and calling it the principal root is legitimate.

Why the product rule fails exactly when both factors are negative

Two examples showed the rule failing whenever both $a$ and $b$ are negative. Here is why that is the only way it fails, for every real $a$ and $b$ .

Use the chapter-5 convention $\sqrt{-c} = i\sqrt c$ for $c > 0$ . If at most one of $a$ and $b$ is negative, say $a = -m$ with $m > 0$ and $b \ge 0$ , then

\sqrt a \, \sqrt b = i\sqrt m \, \sqrt b = i\sqrt{mb}, \qquad \text{and} \qquad \sqrt{ab} = \sqrt{-mb} = i\sqrt{mb},

so the two sides agree. But if both are negative, $a = -m$ and $b = -n$ with $m, n > 0$ , then

\sqrt a \, \sqrt b = (i\sqrt m)(i\sqrt n) = i^2 \sqrt{mn} = -\sqrt{mn}, \qquad \text{while} \qquad \sqrt{ab} = \sqrt{mn} ,

and those two are negatives of each other, never equal (unless both are $0$ , which two negative numbers cannot be). So $\sqrt{ab} = \sqrt a \sqrt b$ holds for every real $a$ and $b$ except when both are negative, and in that one case it fails by exactly a factor of $-1$ .

A bit of history (optional)

Here is a mistake worth more than most correct answers. It was printed in 1770, in an algebra textbook by Leonhard Euler, a Swiss mathematician who is still among the most prolific ever to work in the field.

Euler was using the rule you have been using all lesson. The root of a product is the product of the roots. He applied it to two negative numbers, and concluded that $\sqrt{-2}$ times $\sqrt{-3}$ is $\sqrt6$ . By the conventions now standard the answer is $-\sqrt6$ , the other square root of the same number. Euler had walked straight into the trap this lesson is built around.

It is not hard to see how. In 1770 nobody had yet insisted that the radical sign must name one particular number. The square root of a negative was still handled on instinct rather than by definition. So the rule looked unconditional, because nothing in the notation announced the hypothesis it was quietly resting on.

That is why the clause “provided $a \ge 0$ and $b \ge 0$ ” is not textbook throat clearing. It is the whole difference between the answer you now know how to defend and the one that one of the greatest mathematicians in history wrote down.