Roots and Radicals

A square has area 50 square metres. What is the length of one side?

You probably reach for the answer \sqrt{50} immediately, which is correct — but think about what that symbol is doing. It is a notation for "the number whose square is 50." There is no integer answer (7^2 = 49 and 8^2 = 64, and 50 is between them); there is no terminating decimal answer; there is no neat fraction answer. There is one specific real number, sitting at exactly 7.0710678\dots on the number line, and the symbol \sqrt{50} is its name.

You met a number like this before. In Number Systems, you saw how to construct \sqrt{2} on a number line using a ruler and compass — placing a right-angled triangle with both legs of length 1 and reading off the length of the hypotenuse. The same technique works for \sqrt{50}, and for any positive square root: there is always one specific point on the line that the symbol refers to, and you can pin it down geometrically even when you cannot write it as a decimal. This article is about all such numbers — the square roots, the cube roots, the n-th roots — and the small set of rules that lets you compute with them.

The big idea is that roots are not a separate operation from exponents. They are the same operation, just with the index written underneath the radical sign instead of as a superscript. Once you see the connection, the laws of radicals become the laws of exponents in a different costume, and there is essentially nothing new to memorise.

What a square root is

The square root of a non-negative number a, written \sqrt{a}, is the non-negative number whose square is a. So \sqrt{9} = 3 because 3^2 = 9, and \sqrt{50} \approx 7.071 because 7.071^2 \approx 50.

The qualifier non-negative matters. The equation x^2 = 9 has two solutions, x = 3 and x = -3, since (-3)^2 also equals 9. So in everyday language you might say "the square roots of 9 are \pm 3," and both are correct. But the symbol \sqrt{9} is reserved for the non-negative one — by convention, \sqrt{9} = 3, never -3. This is called the principal square root, and it is what the symbol \sqrt{\,\,} always returns. To get the other root, you write -\sqrt{9} explicitly. The convention exists so that \sqrt{a} is a function — one input, one output — instead of a relation that returns two answers.

You cannot take the square root of a negative number and get a real answer, because no real number squared is negative. \sqrt{-4} has no real value. (It does have a complex value, 2i, where i is the imaginary unit you will meet in a later chapter — but in real-number arithmetic, the symbol is undefined.)

Cube roots and nth roots

The cube root of a number a, written \sqrt[3]{a}, is the number whose cube is a. So \sqrt[3]{8} = 2 because 2^3 = 8, and \sqrt[3]{27} = 3 because 3^3 = 27.

Cube roots are friendlier than square roots in one important way: the cube root of a negative number is a real number. (-2)^3 = -8, so \sqrt[3]{-8} = -2. There is no need for a "principal value" convention here, because the cube of a negative is uniquely negative — there is just one real cube root, and the symbol \sqrt[3]{a} refers to it for any real a, positive or negative.

The general pattern: the n-th root of a, written \sqrt[n]{a}, is the number whose n-th power is a. So \sqrt[4]{16} = 2 because 2^4 = 16, and \sqrt[5]{32} = 2 because 2^5 = 32. The n here is called the index of the radical, and the expression underneath is called the radicand. When the index is 2 — the most common case — it is conventionally omitted: \sqrt{a} means \sqrt[2]{a}.

The behaviour of n-th roots for negative inputs depends on whether n is even or odd:

• If n is odd, \sqrt[n]{a} is defined for every real a (positive, negative, or zero), and there is exactly one real value. The cube root example above is the simplest case.
• If n is even, \sqrt[n]{a} is only defined for a \geq 0, and the convention (like with square roots) is that the result is the non-negative one.

The reason is symmetry: an odd power preserves the sign of its input ((-2)^3 = -8, negative in, negative out), while an even power destroys it ((-2)^2 = 4, negative in, positive out). So even-power equations like x^2 = 4 have two solutions (\pm 2), and even-root symbols pick a single one of them by convention; odd-power equations like x^3 = -8 have only one real solution, so odd-root symbols return it directly.

Roots are exponents in disguise

This is the most important fact about roots, and it is the bridge to everything in Exponents and Powers.

Claim: \sqrt[n]{a} = a^{1/n}, for any positive a and any positive integer n.

Why? Because the laws of exponents leave no other choice. From the power-of-a-power law,

So a^{1/n} is a number whose n-th power is a. And by the definition of \sqrt[n]{\,\,}, that number is \sqrt[n]{a}. The two notations refer to the same thing.

This is the same forced-by-consistency move that gave us a^0 = 1 and a^{-n} = 1/a^n in the previous chapter. The "1/n" exponent isn't a new convention — it is what the exponent laws demand once you decide to allow fractional exponents at all.

The combined rule for general fractional exponents follows immediately:

So 8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4, and equivalently 8^{2/3} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4. Both readings give the same answer, because the laws of exponents make them equivalent.

This identification is the most useful single fact in this article. It means every property of radicals is just a property of exponents in different notation, and you do not have to learn a separate set of rules for radicals once you know the exponent laws. The next section makes this concrete.

Properties of radicals

The five properties below are the working rules for manipulating radical expressions. Each one is one of the laws of exponents from the previous chapter, rewritten with the radical symbol in place of fractional exponents.

Property 1: Product of radicals. For a, b \geq 0:

This is the power-of-a-product law (ab)^{1/2} = a^{1/2} \cdot b^{1/2} written in radical notation. The same rule holds for n-th roots: \sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}.

Property 2: Quotient of radicals. For a \geq 0 and b > 0:

This is the power-of-a-quotient law (a/b)^{1/2} = a^{1/2}/b^{1/2}.

Property 3: Power of a radical. For a \geq 0:

This is just the power-of-a-power law applied to a fractional exponent.

Property 4: Root of a root. For a \geq 0:

A root of a root is a single root with the indices multiplied. In exponent form, \left(a^{1/n}\right)^{1/m} = a^{1/(mn)}, again the power-of-a-power law.

Property 5: A radical of itself. For a \geq 0:

This last one has a small twist: the result is the absolute value of a, not just a. The reason is that the principal square root is always non-negative, but a might be negative — so \sqrt{(-3)^2} = \sqrt{9} = 3 = |-3|, not -3. For non-negative a, the absolute value is the same as a itself, so this twist only matters when a might be negative.

A warning that catches everyone the first time: the product and quotient properties do not extend to addition. There is no rule of the form \sqrt{a + b} = \sqrt{a} + \sqrt{b}. Quick check: \sqrt{9 + 16} = \sqrt{25} = 5, but \sqrt{9} + \sqrt{16} = 3 + 4 = 7. Different. The product-and-quotient laws live inside multiplication; they have no addition analogue, exactly as the distributive law in Operations and Properties goes only one way.

Simplifying radical expressions

Once the properties are in hand, "simplifying" a radical expression means using them to rewrite the expression in a standard cleaner form. The basic moves:

Pull out perfect-square factors. A square root is "simplified" when the radicand has no perfect-square factors other than 1. To simplify \sqrt{50}, factor the radicand and find the largest perfect square inside: 50 = 25 \times 2, and 25 is a perfect square. Then use the product property:

So \sqrt{50} and 5\sqrt{2} are the same number, but 5\sqrt{2} is the standard simplified form. The reader can immediately see that the value is "five point something," because \sqrt{2} \approx 1.414.

The same trick works for n-th roots: pull out perfect n-th powers. To simplify \sqrt[3]{54}, write 54 = 27 \times 2, and 27 = 3^3 is a perfect cube:

Combine like radicals. Two radicals are like if they have the same index and the same radicand after simplification. Like radicals add and subtract the way like terms do in algebra: 3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}, just as 3x + 5x = 8x. Unlike radicals (different radicands or different indices) cannot be combined directly; \sqrt{2} + \sqrt{3} is already as simple as it gets.

Rationalising denominators — revisited

In Fractions and Decimals you saw the trick for cleaning up a fraction whose denominator contains a square root. The same trick is even more useful for radical expressions, so here is a quick refresher and an extension.

Single radical denominator. Multiply top and bottom by the same radical:

The denominator becomes \sqrt{3} \cdot \sqrt{3} = (\sqrt{3})^2 = 3, a clean integer.

Binomial denominator with a radical. Use the conjugate trick. The conjugate of a + b\sqrt{c} is a - b\sqrt{c}, and their product is the rational number a^2 - b^2 c (no radicals left). Example:

The reason rationalisation is worth doing: the standard form makes comparison easier (which fraction is bigger?), makes adding fractions with radical denominators tractable (you need a common denominator, and the LCM of \sqrt{3} and \sqrt{2} is meaningless until both have rational denominators), and historically made manual computation possible (long division by an irrational number is impractical, but long division by an integer is mechanical).

Two worked examples

Common confusions

• "\sqrt{a + b} = \sqrt{a} + \sqrt{b}." No. Quick check: \sqrt{9 + 16} = \sqrt{25} = 5, but \sqrt{9} + \sqrt{16} = 3 + 4 = 7. The product and quotient properties do not extend to addition. There is no clean rule for \sqrt{a + b}.

• "\sqrt{9} is \pm 3." As an equation solver, x^2 = 9 does have two solutions \pm 3. But the symbol \sqrt{9} refers to the principal (non-negative) square root, which is just 3. To get the negative one, you write -\sqrt{9} explicitly.

• "\sqrt{a^2} = a." Almost — the correct identity is \sqrt{a^2} = |a|. If a might be negative, the absolute value is necessary. This shows up most often when solving equations like x^2 = 16, where the answer is |x| = 4, that is, x = \pm 4.

• "You can't take the cube root of a negative number." You can. (-2)^3 = -8, so \sqrt[3]{-8} = -2. The "no negative inputs" rule only applies to even roots (square roots, fourth roots, sixth roots, …), because no real number raised to an even power can be negative. Odd roots are unrestricted.

• "8^{2/3} means 8 \div (2/3) or something weird." No — 8^{2/3} is a power, and the laws of exponents say 8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4. The fraction in the exponent is read literally as a fractional exponent, and the rule "a^{m/n} = \sqrt[n]{a^m}" tells you what it means.

• "Rationalising the denominator changes the value." No — rationalising multiplies top and bottom by the same thing, which is multiplying by 1. The value is unchanged; only the form of the expression is different. (Same warning that appeared in Fractions and Decimals; it is worth repeating because it is the most common reason students don't trust the technique.)

Going deeper

Most square roots are irrational

The square root of any positive integer that is not a perfect square is irrational. So \sqrt{1}, \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}, \dots are all integers, but the gaps between them — \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \dots — are all irrational. The same is true for n-th roots of any non-perfect n-th power: \sqrt[3]{2} is irrational, \sqrt[3]{3} is irrational, \sqrt[5]{17} is irrational, and so on.

This is a striking statement: irrationality is the normal state for roots, and rationality is the rare exception. The reason is a beautiful proof by contradiction that you may meet in a chapter on number theory, but the intuition is short: if \sqrt{p} for a prime p were rational, you would be able to write it as a/b in lowest terms, and squaring would give p \cdot b^2 = a^2. The factor of p on the left would force p to also be a factor of a^2, hence of a, which would in turn force p to be a factor of b — contradicting "lowest terms." The same argument works for any prime, and a more careful version handles arbitrary non-perfect-square integers.

The geometric story: every square root is constructible

In Number Systems, you saw \sqrt{2} constructed on the number line using a ruler and compass. The same construction generalises: given any positive number a on the line, you can construct \sqrt{a} using only ruler and compass. The recipe is to draw a semicircle whose diameter is a + 1, then drop a perpendicular from the boundary at the point where the a segment meets the 1 segment — the height of the perpendicular is exactly \sqrt{a}. This is one of the oldest theorems in geometry and gives a satisfying answer to "where does the number \sqrt{a} live?" — it lives at the height of a specific perpendicular in a specific semicircle.

The constructibility of square roots is also why straightedge-and-compass constructions can produce all the lengths that show up in classical Greek geometry, but cannot in general produce cube roots — and that limitation is the reason "doubling the cube" (constructing a cube of exactly twice the volume of a given one, which requires constructing \sqrt[3]{2}) was famously impossible with ruler and compass alone, a result not proved until the nineteenth century.

The conjugate trick is the same trick for complex numbers

The conjugate of \sqrt{5} - \sqrt{3} is \sqrt{5} + \sqrt{3}, and their product is rational. The conjugate of the complex number a + bi (where i is the imaginary unit) is a - bi, and their product is a^2 + b^2, a real number with no imaginary part. The two tricks are the same trick, applied to two different "kinds of irrationality" — radicals in the first case, imaginary numbers in the second.

This is one of the clean connections that the field of algebra makes possible: the same algebraic move (multiply by the conjugate, use the difference-of-squares pattern) does the same kind of work in two completely different settings. The reason it works in both places is that both irrationals and imaginaries satisfy a quadratic equation with rational coefficients — and the conjugate is the other root of that quadratic. It is the same idea wearing two different costumes.

Where this leads next

Roots are the bridge between school arithmetic and the chapter on quadratic equations, where they appear in the answer to almost every problem.

• Exponents and Powers — the previous chapter, which contains the laws that every property of radicals on this page secretly is.
• Fractions and Decimals — where the rationalising-denominators trick first appeared, in a slightly less general form.
• Quadratic Equations — every quadratic formula answer involves a square root, and the rationalising-the-denominator trick reappears constantly in simplifying those answers.
• Number Systems — the picture of where roots like \sqrt{2} and \sqrt[3]{2} live on the real line, and the geometric construction of \sqrt{a}.
• Operations and Properties — the structural rules behind every algebraic manipulation in this article.