Operations with Radicals

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 equaled $\sqrt{a+b}$, then taking $a = 9$ and $b = 16$ would give:

Seven is not five, so the rule is false. That one pair of perfect squares is enough to prove it.

Now look at what does work:

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:

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 only decides the question once both radicals are already 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.

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.

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:

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

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 can mislead you: $\sqrt{12}$ and $\sqrt{27}$ look unlike and are nonetheless the same unit five times over. Simplifying is not cosmetic housekeeping before addition. It is the step that tells you whether addition is possible at all.

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.

Multiplying radicals

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

Multiplying two radical terms uses that rule twice: multiply the coefficients, multiply the radicands, then simplify what comes out. A product of two unsimplifiable radicals can still hide a square.

In general, because multiplication is commutative and associative,

That hypothesis $a, b \ge 0$ 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 in the complex numbers, where $\sqrt{-4} = 2i$ and $\sqrt{-9} = 3i$, the rule collapses:

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

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 multiplying a radical against a sum works like multiplying anything else against a sum: distribute it across each term.

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

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

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 $(3 + 2\sqrt5)(3 - 2\sqrt5) = 9 - 4 \cdot 5 = -11$. You will use this same move in the chapter’s last lesson to clear a radical out of a denominator. For now, just register the fact: pairing a radical sum with its conjugate destroys the radical.

Dividing radicals

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

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:

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 an earlier lesson’s translation into rational exponents, where a root of index $n$ becomes an exponent of $1/n$:

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:

This is the payoff of that earlier lesson. Fractional exponents are not just a different way to write a root you already knew. They are what turns a question about incompatible radical signs into a question about adding fractions, which you already know how to do.

Common mistakes