Roots

A root is the inverse of raising to a power. The square root of a number $a \ge 0$ is the non-negative number $b$ with $b^2 = a$:

$$\sqrt{a} = b \iff b^2 = a \quad (a, b \ge 0)$$

For example $\sqrt{49} = 7$, because $7^2 = 49$.

The condition $a \ge 0$ is not incidental: the square of every real number is non-negative, so a negative number has no square root among the real numbers.

Cube roots behave differently:

$$\sqrt[3]{a} = b \iff b^3 = a$$

Here a negative input is fine, because the cube of a negative number is negative: $\sqrt[3]{-8} = -2$.

Arithmetic with roots

Roots distribute over multiplication and division:

$$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$

But not over addition: $\sqrt{9 + 16} = \sqrt{25} = 5$, while $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$.

Taking a factor out of a radical

Most roots are not rational numbers — $\sqrt{72}$ cannot be written as a fraction. It can, however, be simplified: spot a square factor under the radical and take it out:

$$\sqrt{a^2 \cdot b} = a\sqrt{b} \quad (a \ge 0)$$

Rationalising a denominator

A fraction with a root in the denominator is usually rewritten by multiplying the numerator and the denominator by that same root — that is, by one:

$$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{\sqrt{a}}{a} \quad (a > 0)$$

Hence the familiar $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ — the same number, a handier form.

A root is a power

A root is simply a power with a fractional exponent:

$$\sqrt{a} = a^{\frac{1}{2}} \qquad \sqrt[n]{a} = a^{\frac{1}{n}}$$

That ties the two topics together: every law of exponents then holds for roots as well, with no separate rules to memorise.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

√121 =ResultCheck

Common mistakes

• Splitting a root over addition — $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$.
• Writing $\sqrt{16}$ as $\pm 4$ — a square root is non-negative by definition; the $\pm$ appears only when solving the equation $x^2 = 16$.
• Stopping at the approximation — when a question asks for the exact value, the answer is $6\sqrt{2}$, not $8.49$.
• Looking for the square root of a negative number — it does not exist among the reals, unlike the cube root.