Operations with Radicals

Learning goals

Combine like radicals by the distributive law
Simplify before deciding whether two radicals are alike
Multiply coefficients with coefficients and radicands with radicands
Use a conjugate pair to erase the radicals
Switch to rational exponents when the indices differ

Radicals add like terms, not like radicands

Start by killing the tempting rule with numbers you can check by hand. If $\sqrt a + \sqrt b$ really equalled $\sqrt{a+b}$ , then taking $a = 9$ and $b = 16$ would give:

\sqrt9 + \sqrt{16} = 3 + 4 = 7, \qquad \text{but} \qquad \sqrt{9 + 16} = \sqrt{25} = 5.

Seven is not five, so the rule is false. Notice how little effort that refutation took: one pair of perfect squares, and the claim is gone.

Now look at what does work:

\sqrt2 + \sqrt2 = (1 + 1)\sqrt2 = 2\sqrt2.

Read the middle step carefully, because it is the whole lesson. Nothing about square roots was used there. That is the distributive law, $cu + du = (c + d)u$ , applied with $u = \sqrt2$ . The same law that turns $3x + 5x$ into $8x$ turns $3\sqrt2 + 5\sqrt2$ into $8\sqrt2$ , and for the same reason:

3\sqrt2 + 5\sqrt2 = (3 + 5)\sqrt2 = 8\sqrt2.

Two radicals are like radicals when they have the same index and the same radicand. The index and the radicand together name the unit; the number in front is how many of that unit you have. That test decides whether two terms combine only once both radicals are in simplest form, a restriction the next section shows is doing real work. So $\sqrt2$ and $4\sqrt2$ are like radicals, while $\sqrt2$ and $\sqrt3$ are not (different radicands). And $\sqrt2$ and $\sqrt[3]{2}$ are not like radicals either (same radicand, different indices, so they are not even the same size: $\sqrt2 \approx 1.414$ and $\sqrt[3]{2} \approx 1.260$ ).

Only like radicals collect, because only like radicals share the common factor that the distributive law needs. There is no rule for adding $\sqrt2$ and $\sqrt3$ for the same reason there is no rule for adding $x$ and $y$ . That is because the sum $x + y$ is already as simple as it gets.

Laying the two lengths end to end reaches about 3.146, while the square root of 5 stops at about 2.236. The sum of the roots is not the root of the sum.

The picture shows that the two quantities differ, but it does not say by how much, or whether the false rule might rescue itself in some lucky case. Algebra settles both questions at once.

$\sqrt a + \sqrt b = \sqrt{a+b}$ holds only when $a = 0$ or $b = 0$ #

Let $a \ge 0$ and $b \ge 0$ , so that all three square roots are real, and suppose $\sqrt a + \sqrt b = \sqrt{a + b}$ . Equal numbers have equal squares, so square both sides. Expanding the left side with the distributive law gives $(\sqrt a + \sqrt b)^2 = (\sqrt a)^2 + 2\sqrt a \sqrt b + (\sqrt b)^2$ , and since $a$ and $b$ are nonnegative we may use the product rule $\sqrt a \sqrt b = \sqrt{ab}$ , so the left side becomes $a + b + 2\sqrt{ab}$ . The right side squares to $a + b$ . The supposed identity therefore says

a + b + 2\sqrt{ab} = a + b,

and cancelling $a + b$ leaves $2\sqrt{ab} = 0$ , hence $\sqrt{ab} = 0$ , hence $ab = 0$ . A product of two numbers is zero only when one of them is zero, so $a = 0$ or $b = 0$ .

The converse is immediate. If $a = 0$ , the claim reads $\sqrt0 + \sqrt b = \sqrt{0 + b}$ , which is just $\sqrt b = \sqrt b$ , and the case $b = 0$ is identical. So the identity holds for exactly those pairs and no others.

The proof does more than refute the rule; it shows where the rule leaks. Squaring the two sides gives $(\sqrt a + \sqrt b)^2 = (a + b) + 2\sqrt{ab}$ against $(\sqrt{a + b})^2 = a + b$ , so the two squares differ by exactly $2\sqrt{ab}$ . For positive $a$ and $b$ that cross term is positive, and squaring is strictly increasing on nonnegative numbers, so $\sqrt a + \sqrt b$ is strictly the larger of the two. That cross term is what a student who writes $\sqrt a + \sqrt b = \sqrt{a+b}$ is quietly throwing away.

Be careful about what $2\sqrt{ab}$ measures: it is the gap between the squares, not the gap between the numbers themselves. At $a = 9$ and $b = 16$ the two sides are $7$ and $5$ , a gap of $2$ , while $2\sqrt{ab} = 24$ is the gap between $49$ and $25$ .

Simplify first, then decide what is alike

Here is the trap that makes the like-radical test harder than it looks. Consider $\sqrt{12} + \sqrt{27}$ . The radicands are $12$ and $27$ , which are different, so the two radicals are unlike and the sum cannot be collected. That conclusion is wrong.

Simplify each radical first, pulling out the largest perfect square:

\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt3, \qquad \sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt3.

They were like radicals all along, wearing a disguise. Now the distributive law finishes the job:

\sqrt{12} + \sqrt{27} = 2\sqrt3 + 3\sqrt3 = (2 + 3)\sqrt3 = 5\sqrt3.

So state the test carefully. Put every radical in simplest form, and only then compare. Two radical terms in simplest form combine into a single term exactly when they have the same index and the same radicand. Applied to radicals you have not yet simplified, that test is worthless: $\sqrt{12}$ and $\sqrt{27}$ pass the eye test for “unlike” and are nonetheless the same unit five times over. This is why simplifying is not cosmetic housekeeping before addition. It is the step that tells you whether addition is possible at all.

Both radicals are built from the same unit. Two copies of it make the square root of 12, three copies make the square root of 27, and together they are five copies.

The index has to match as well, and simplifying can expose a shared index just as easily. Cube roots collect among themselves: since $\sqrt[3]{16} = \sqrt[3]{8 \cdot 2} = 2\sqrt[3]{2}$ and $\sqrt[3]{54} = \sqrt[3]{27 \cdot 2} = 3\sqrt[3]{2}$ , we get $\sqrt[3]{16} + \sqrt[3]{54} = 5\sqrt[3]{2}$ . But $\sqrt2 + \sqrt[3]{2}$ stays exactly as written. The radicands agree and the indices do not, so there is no common unit to factor out.

Worked example 1 Simplify $\sqrt{50} + \sqrt{18} - \sqrt{8}$

None of the three radicands is a perfect square, and all three are different, so nothing can be decided yet. Simplify each one by splitting off its largest square factor.

\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt2, \qquad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt2, \qquad \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt2.

Every term turned out to be a multiple of the same unit $\sqrt2$ , so the distributive law applies to all three at once:

5\sqrt2 + 3\sqrt2 - 2\sqrt2 = (5 + 3 - 2)\sqrt2 = 6\sqrt2.

As a sanity check, $\sqrt{50} + \sqrt{18} - \sqrt{8} \approx 7.071 + 4.243 - 2.828 = 8.486$ , and $6\sqrt2 \approx 8.485$ . The two agree, as they must.

Check your understanding

Simplify $\sqrt{45} + \sqrt{20}$ .

\sqrt{65}

5\sqrt{10}

5\sqrt5

\sqrt5

Multiplying radicals

Multiplication is where radicals are cooperative, and the reason is the product rule from the last lesson: for $a \ge 0$ and $b \ge 0$ ,

\sqrt a \cdot \sqrt b = \sqrt{ab}.

Carry that hypothesis with you; it is not decoration. Take $a = -4$ and $b = -9$ . Over the real numbers $\sqrt{-4}$ does not exist, so there is nothing to multiply. Even after the complex numbers give $\sqrt{-4} = 2i$ and $\sqrt{-9} = 3i$ , the rule collapses:

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

The two answers differ in sign, so $a, b \ge 0$ is what keeps the product rule honest.

With that in hand, a product of two radical terms splits into two independent products, because multiplication is commutative and associative:

(c\sqrt a)(d\sqrt b) = (c \cdot d)(\sqrt a \cdot \sqrt b) = cd\sqrt{ab} \qquad (a, b \ge 0).

Coefficients multiply with coefficients, radicands with radicands. Then simplify what comes out, since a product of two unsimplifiable radicals can easily hide a square:

(2\sqrt6)(3\sqrt{10}) = 6\sqrt{60} = 6\sqrt{4 \cdot 15} = 6 \cdot 2\sqrt{15} = 12\sqrt{15}.

Two special cases fall straight out. Squaring undoes the root, $(\sqrt a)^2 = a$ for $a \ge 0$ , because that is exactly what the principal square root is defined to do, so $(5\sqrt3)^2 = 25 \cdot 3 = 75$ . And a radical distributes over a sum like any other factor:

\sqrt3(\sqrt6 + \sqrt{15}) = \sqrt{18} + \sqrt{45} = 3\sqrt2 + 3\sqrt5.

Products of two binomials work exactly as they did for polynomials: expand every term against every term, then collect like radicals.

(2 + \sqrt3)(5 - \sqrt3) = 10 - 2\sqrt3 + 5\sqrt3 - (\sqrt3)^2 = 10 + 3\sqrt3 - 3 = 7 + 3\sqrt3.

Now one product deserves to be singled out. Multiply a sum of two radicals by the difference of the same two radicals, and the middle terms cancel by the difference-of-squares pattern:

(\sqrt a + \sqrt b)(\sqrt a - \sqrt b) = (\sqrt a)^2 - (\sqrt b)^2 = a - b \qquad (a, b \ge 0).

Look at what that says. Two irrational numbers can multiply to a plain integer: $(\sqrt7 + \sqrt3)(\sqrt7 - \sqrt3) = 7 - 3 = 4$ . Every radical is gone. The same works with a rational term attached, $(p + q\sqrt c)(p - q\sqrt c) = p^2 - q^2 c$ , so $(3 + 2\sqrt5)(3 - 2\sqrt5) = 9 - 4 \cdot 5 = -11$ . This is the single most useful product in the whole chapter. That product is also the engine behind the last lesson, where you will use it to clear radicals out of a denominator. Here, just register the fact: pairing a radical sum with its difference destroys the radical.

Worked example 2 Expand $(2\sqrt3 + \sqrt5)(\sqrt3 - 3\sqrt5)$

Multiply each term of the first factor against each term of the second, keeping the coefficients and the radicands in separate columns.

In two of the four products a radical meets its twin and vanishes: $2\sqrt3 \cdot \sqrt3 = 2 \cdot 3 = 6$ and $\sqrt5 \cdot (-3\sqrt5) = -3 \cdot 5 = -15$ . The two mixed products keep their radicals: $2\sqrt3 \cdot (-3\sqrt5) = -6\sqrt{15}$ and $\sqrt5 \cdot \sqrt3 = \sqrt{15}$ .

Assemble the four pieces and collect the like radicals, which here are the two $\sqrt{15}$ terms:

6 - 6\sqrt{15} + \sqrt{15} - 15 = (6 - 15) + (-6 + 1)\sqrt{15} = -9 - 5\sqrt{15}.

Numerically the original product is about $5.700 \times (-4.976) \approx -28.36$ , and $-9 - 5\sqrt{15} \approx -9 - 19.36 = -28.36$ , so the expansion holds up.

Worked example 3 Compare $(\sqrt{11} + \sqrt6)(\sqrt{11} - \sqrt6)$ with $(\sqrt{11} + \sqrt6)^2$

These two look almost the same, and one sign makes all the difference.

For the first, the cross terms are $-\sqrt{11}\sqrt6$ and $+\sqrt6\sqrt{11}$ , which are negatives of each other and cancel:

(\sqrt{11} + \sqrt6)(\sqrt{11} - \sqrt6) = 11 - 6 = 5.

For the second, the cross terms have the same sign, so instead of cancelling they double:

(\sqrt{11} + \sqrt6)^2 = 11 + 2\sqrt{11}\sqrt6 + 6 = 17 + 2\sqrt{66}.

The first is an integer; the second is not, and it is nowhere near $11 + 6 = 17$ , because $2\sqrt{66} \approx 16.25$ is a large piece of the answer. Squaring a sum never lets you square the terms and stop, whether the terms are radicals or anything else.

Check your understanding

Expand $(\sqrt{10} - \sqrt3)(\sqrt{10} + \sqrt3)$ .

\sqrt7

13

7 + 2\sqrt{30}

7

Dividing radicals

Division follows the same pattern with the same kind of hypothesis. For $a \ge 0$ and $b > 0$ ,

\frac{\sqrt a}{\sqrt b} = \sqrt{\frac{a}{b}}.

The reason is short. The quotient $\sqrt a / \sqrt b$ is nonnegative, and squaring it gives $(\sqrt a)^2 / (\sqrt b)^2 = a / b$ . The principal square root of $a/b$ is by definition the one nonnegative number whose square is $a/b$ , so the quotient must be it. We need $b > 0$ rather than $b \ge 0$ only to keep from dividing by zero.

In practice this turns ugly-looking quotients into clean ones:

\frac{\sqrt{72}}{\sqrt2} = \sqrt{36} = 6, \qquad \frac{12\sqrt{21}}{3\sqrt7} = 4\sqrt{\frac{21}{7}} = 4\sqrt3.

When the division does not come out clean, as with $\sqrt2 / \sqrt3$ , you are left with a radical sitting in the denominator. That is not wrong, merely unfinished, and the chapter’s last lesson is devoted to tidying it up.

Roots of different indices

Now the case where the product rule is genuinely powerless. What is $\sqrt2 \cdot \sqrt[3]{2}$ ?

The product rule needs a common index, and here there is none: one factor is a square root, the other a cube root. No amount of staring will fix that, because the rule simply does not apply. What does apply is the previous lesson’s translation into rational exponents, where a root of index $n$ becomes an exponent of $1/n$ :

\sqrt2 \cdot \sqrt[3]{2} = 2^{1/2} \cdot 2^{1/3} = 2^{1/2 + 1/3} = 2^{5/6} = \sqrt[6]{2^5} = \sqrt[6]{32}.

Powers of the same base multiply by adding exponents, and adding $1/2$ and $1/3$ forces you to find a common denominator. That common denominator, $6$ , is the common index the radicals were missing. The same computation in radical clothing says: rewrite $\sqrt2 = 2^{3/6} = \sqrt[6]{8}$ and $\sqrt[3]{2} = 2^{2/6} = \sqrt[6]{4}$ , and now that the indices agree the product rule finally applies, giving $\sqrt[6]{8} \cdot \sqrt[6]{4} = \sqrt[6]{32}$ . Same answer, and it shows what “common index” means on the page.

The radicands do not even have to be powers of one base. The common-index idea still carries it:

\sqrt2 \cdot \sqrt[3]{3} = 2^{3/6} \cdot 3^{2/6} = \sqrt[6]{2^3} \cdot \sqrt[6]{3^2} = \sqrt[6]{8 \cdot 9} = \sqrt[6]{72}.

This is the payoff of the previous lesson. Fractional exponents are not a notational curiosity that dresses up roots you already knew how to write. They are what makes mixed-index arithmetic possible at all. They do it by turning a question about incompatible radical signs into a question about adding fractions, which you have been able to do for years.

Worked example 4 Write $\sqrt[3]{9}\cdot\sqrt{3}$ in simplest form

The indices are $3$ and $2$ , so the product rule is unavailable. Convert both factors to powers of the same base. Since $9 = 3^2$ ,

\sqrt[3]{9} = (3^2)^{1/3} = 3^{2/3}, \qquad \sqrt{3} = 3^{1/2}.

Multiplying powers of $3$ means adding the exponents, over the common denominator $6$ :

3^{2/3} \cdot 3^{1/2} = 3^{2/3 + 1/2} = 3^{4/6 + 3/6} = 3^{7/6}.

An exponent bigger than $1$ means a whole factor of $3$ can be pulled out, since $7/6 = 1 + 1/6$ :

3^{7/6} = 3^1 \cdot 3^{1/6} = 3\sqrt[6]{3}.

Checking numerically, $\sqrt[3]{9} \cdot \sqrt3 \approx 2.080 \times 1.732 \approx 3.603$ , and $3\sqrt[6]{3} \approx 3 \times 1.201 \approx 3.603$ .

Check your understanding

Simplify $\sqrt[3]{2}\cdot\sqrt[6]{2}$ .

\sqrt[9]{2}

\sqrt2

\sqrt[18]{2}

2

Common mistakes

Writing $\sqrt a + \sqrt b = \sqrt{a + b}$ . The fastest refutation is a pair of perfect squares: $\sqrt9 + \sqrt{16} = 3 + 4 = 7$ , but $\sqrt{9 + 16} = \sqrt{25} = 5$ . The rule holds only when $a$ or $b$ is $0$ , and never otherwise.
Declaring radicals unlike before simplifying them. $\sqrt{12}$ and $\sqrt{27}$ have different radicands and nonetheless collect, because in simplest form they are $2\sqrt3$ and $3\sqrt3$ . They are not like radicals as written; their simplest forms are. Simplify first, because the like-radical test only means anything in simplest form.
Combining the radicands of like radicals. $3\sqrt5 + 4\sqrt5$ is $7\sqrt5$ , not $7\sqrt{10}$ and not $7\sqrt{25}$ . The radicand is the unit, and the unit does not change when you count more of it, exactly as $3x + 4x = 7x$ and not $7x^2$ .
Ignoring the index. $\sqrt2$ and $\sqrt[3]{2}$ share a radicand but are different numbers ( $1.414$ against $1.260$ ), so they are not like radicals and cannot be collected. To multiply them, go through rational exponents.
Using the product rule on negative radicands. $\sqrt{-4} \cdot \sqrt{-9}$ is not $\sqrt{36} = 6$ . In the complex numbers it is $(2i)(3i) = -6$ . The hypothesis $a, b \ge 0$ is doing real work.
Squaring a sum term by term. $(\sqrt a + \sqrt b)^2$ is $a + 2\sqrt{ab} + b$ , not $a + b$ . The cross term is the same $2\sqrt{ab}$ that makes $\sqrt a + \sqrt b = \sqrt{a+b}$ false, so this is the first mistake in disguise.

Key takeaways

Like radicals have the same index and the same radicand. They combine by the distributive law, $c\sqrt n + d\sqrt n = (c + d)\sqrt n$ , exactly as like terms do, with $\sqrt n$ playing the part of the variable. Two radical terms in simplest form combine exactly when they are like radicals, and unlike radicals in simplest form do not combine at all.
Simplify before you compare. $\sqrt{12} + \sqrt{27}$ becomes $2\sqrt3 + 3\sqrt3 = 5\sqrt3$ . Radicals that look unlike can share a unit once the square factors are pulled out, which is why the test above is worthless until you have simplified.
$\sqrt a + \sqrt b = \sqrt{a + b}$ is false for every positive $a$ and $b$ . Squaring shows why: the two squares differ by exactly the cross term $2\sqrt{ab}$ , so the left side is strictly the larger.
Product and quotient rules need nonnegative radicands: $\sqrt a \sqrt b = \sqrt{ab}$ for $a, b \ge 0$ , and $\sqrt a / \sqrt b = \sqrt{a/b}$ for $a \ge 0$ , $b > 0$ . Multiply coefficients with coefficients and radicands with radicands, then simplify.
Conjugate pairs erase radicals: $(\sqrt a + \sqrt b)(\sqrt a - \sqrt b) = a - b$ , and $(p + q\sqrt c)(p - q\sqrt c) = p^2 - q^2c$ . Squaring the same sum does the opposite, doubling the cross term into $2\sqrt{ab}$ .
Different indices need rational exponents. The product rule cannot touch $\sqrt2 \cdot \sqrt[3]{2}$ , but $2^{1/2} \cdot 2^{1/3} = 2^{5/6} = \sqrt[6]{32}$ settles it. The common denominator of the exponents is the common index the radicals lacked.

Operations with Radicals

Learning goals

Radicals add like terms, not like radicands

Simplify first, then decide what is alike

Multiplying radicals

Dividing radicals

Roots of different indices

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)