Simplifying Radical Expressions

One symbol, one number

Start with a question you can already answer. The equation

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. Notation has to be a function: one input, one output.

So a choice was made, and it is the foundation of everything below.

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:

• $\sqrt a$ is never negative. Whatever else an expression $\sqrt{\text{stuff}}$ turns out to equal, 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:

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 it should be earned rather than assumed.

Roots of any index

Everything above generalises. 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:

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:

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 is stated over the real numbers with a non-negative radicand. Hold onto that, because the next section is where ignoring it 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 $x = -3$:

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.

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

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:

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).

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:

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.

Read the last two paragraphs again. 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:

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

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:

Off by a sign again, and always by a sign, because the two candidate roots differ only by a sign. 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.

Sorting the cases out gives a sharp statement. For real $a$ and $b$, use the chapter-5 convention $\sqrt{-c} = i\sqrt c$ for $c > 0$. Then the identity $\sqrt{ab} = \sqrt a \sqrt b$ holds unless $a$ and $b$ are both negative, and when they are both negative it always fails, by a factor of $-1$. Both directions of that claim are quick to check. If at most one of them is negative, say $a = -m$ with $m > 0$ and $b \ge 0$, then $\sqrt a \sqrt b = i\sqrt m \sqrt b = i\sqrt{mb}$ and $\sqrt{ab} = \sqrt{-mb} = i\sqrt{mb}$, which agree. If both are negative, $a = -m$ and $b = -n$ with $m, n > 0$, then

and those two are negatives of each other, never equal (unless both are zero, which two negatives cannot be).

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.

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 is what the very first proof established for every $n$. Replace each square root by an $n$-th root and the argument runs word for word, so the rule holds at every index:

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 $a/b$:

The condition $b > 0$ is stronger than $b \ge 0$ only because you cannot divide by zero. The same statement holds at any index, for a non-negative $a$ and a positive $b$:

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:

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.

1. No perfect $n$-th power factors remain under the radical. $\sqrt{72}$ fails this, because $36$ divides $72$.
2. No radical appears in a denominator. $\dfrac{1}{\sqrt 2}$ fails this.
3. 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.

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.

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.

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.

Common mistakes