The laws of surds

Surds fol­low a small set of rules that let you mul­ti­ply and divide them, often turn­ing two surds into a whole num­ber. This les­son states the six laws, shows each one with an exam­ple, and then uses them to mul­ti­ply surd expres­sions with brack­ets, includ­ing a pair of con­ju­gates.

The six laws#

In the laws below, $a$ and $b$ are pos­i­tive num­bers and $n$ is a nat­ural num­ber greater than $1$.

Law 1. The prod­uct of two square roots is the square root of the prod­uct:

$$\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$$

For exam­ple, $\sqrt{3} \times \sqrt{5} = \sqrt{3 \times 5} = \sqrt{15}$.

Law 2. A square root times itself gives back the num­ber:

$$\sqrt{a} \times \sqrt{a} = (\sqrt{a})^{2} = \sqrt{a^{2}} = a$$

For exam­ple, $\sqrt{5} \times \sqrt{5} = (\sqrt{5})^{2} = \sqrt{5^{2}} = 5$. This is the law that makes surds dis­ap­pear, and it is used in almost every ques­tion in this chap­ter.

Law 3. The quo­tient of two square roots is the square root of the quo­tient:

$$\displaystyle \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$

For exam­ple, $\displaystyle \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}}$.

Law 4. A root can be writ­ten as a power with a frac­tion for its index:

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

For exam­ple, $\displaystyle \sqrt[3]{2} = 2^{\frac{1}{3}}$. You will meet this idea again, in more depth, in the course on expo­nents.

Law 5. Laws 1 and 3 hold for roots of any order, as long as both roots have the same order:

$$\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{a \times b}$$

For exam­ple, $\sqrt[3]{2} \times \sqrt[3]{3} = \sqrt[3]{2 \times 3} = \sqrt[3]{6}$.

Law 6.

$$\displaystyle \frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}$$

For exam­ple, $\displaystyle \frac{\sqrt[3]{3}}{\sqrt[3]{4}} = \sqrt[3]{\frac{3}{4}}$.

Two things the laws do not say#

Roots do not add like this. There is no law for $\sqrt{a} + \sqrt{b}$. For exam­ple, $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$, but $\sqrt{9 + 16} = \sqrt{25} = 5$. So $\sqrt{a} + \sqrt{b}$ is not $\sqrt{a + b}$.

Dif­fer­ent orders do not com­bine. Laws 5 and 6 need the same order. $\sqrt{2} \times \sqrt[3]{3}$ can­not be writ­ten as one root by these laws.

Using the laws#

Exam­ple 1#

Sim­plify $\sqrt[5]{16} \times \sqrt[5]{2}$.

Both roots have order $5$, so use Law 5.

$$\begin{aligned} \sqrt[5]{16} \times \sqrt[5]{2} &= \sqrt[5]{16 \times 2} \\ &= \sqrt[5]{32} \\ &= \sqrt[5]{2^{5}} = 2 \end{aligned}$$

Two surds mul­ti­plied to give a whole num­ber.

Exam­ple 2#

Sim­plify $\displaystyle \frac{\sqrt[4]{243}}{\sqrt[4]{3}}$.

$$\displaystyle \begin{aligned} \frac{\sqrt[4]{243}}{\sqrt[4]{3}} &= \sqrt[4]{\frac{243}{3}} \\ &= \sqrt[4]{81} \\ &= \sqrt[4]{3^{4}} = 3 \end{aligned}$$

Exam­ple 3#

Sim­plify $(4\sqrt{3})^{2}$.

Square each part of the prod­uct sep­a­rately, then use Law 2.

$$\begin{aligned} (4\sqrt{3})^{2} &= (4)^{2} \times (\sqrt{3})^{2} \\ &= 16 \times 3 \\ &= 48 \end{aligned}$$

Mul­ti­ply­ing brack­ets that con­tain surds#

Brack­ets with surds are mul­ti­plied exactly like brack­ets in alge­bra: mul­ti­ply every term in the first bracket by every term in the sec­ond, then col­lect like terms. Treat $\sqrt{3}$ like a let­ter, except that $\sqrt{3} \times \sqrt{3} = 3$.

Exam­ple 4: con­ju­gates#

Sim­plify $(2 + \sqrt{3})(2 - \sqrt{3})$.

$$\begin{aligned} (2 + \sqrt{3})(2 - \sqrt{3}) &= 4 - 2\sqrt{3} + 2\sqrt{3} - 3 \\ &= 4 - 3 \\ &= 1 \end{aligned}$$

The two surd terms, $-2\sqrt{3}$ and $+2\sqrt{3}$, can­cel. That is what always hap­pens when you mul­ti­ply con­ju­gate surds, and it is why they are so use­ful.

Exam­ple 5#

Sim­plify $(2 + \sqrt{3})(3 - \sqrt{3})$.

$$\begin{aligned} (2 + \sqrt{3})(3 - \sqrt{3}) &= 6 - 2\sqrt{3} + 3\sqrt{3} - 3 \\ &= 3 + \sqrt{3} \end{aligned}$$

Here the brack­ets are not con­ju­gates, so the surd terms do not can­cel: $-2\sqrt{3} + 3\sqrt{3} = \sqrt{3}$. The answer has a ratio­nal part, $3$, and a surd part, $\sqrt{3}$, just as $3 + x$ has a num­ber part and a let­ter part.

Exam­ple 6: squar­ing a bracket#

Sim­plify $(2 + \sqrt{3})^{2}$.

$$\begin{aligned} (2 + \sqrt{3})^{2} &= (2 + \sqrt{3})(2 + \sqrt{3}) \\ &= 4 + 2\sqrt{3} + 2\sqrt{3} + 3 \\ &= 7 + 4\sqrt{3} \end{aligned}$$

Exam­ple 7#

Sim­plify $(\sqrt{3} - \sqrt{2})^{2}$.

$$\begin{aligned} (\sqrt{3} - \sqrt{2})^{2} &= (\sqrt{3} - \sqrt{2})(\sqrt{3} - \sqrt{2}) \\ &= 3 - \sqrt{6} - \sqrt{6} + 2 \\ &= 5 - 2\sqrt{6} \end{aligned}$$

By Law 1, $\sqrt{3} \times \sqrt{2} = \sqrt{6}$.

Prac­tice#

1. Sim­plify $\sqrt{6} \times \sqrt{7}$.
2. Sim­plify $\sqrt{12} \times \sqrt{3}$.
3. Sim­plify $\displaystyle \frac{\sqrt{50}}{\sqrt{2}}$.
4. Sim­plify $\sqrt[3]{4} \times \sqrt[3]{16}$.
5. Sim­plify $\displaystyle \frac{\sqrt[4]{32}}{\sqrt[4]{2}}$.
6. Sim­plify $\sqrt[3]{5} \times \sqrt[3]{25}$.
7. Sim­plify $(3\sqrt{2})^{2}$.
8. Sim­plify $(5 + \sqrt{2})(5 - \sqrt{2})$.
9. Sim­plify $(3 + \sqrt{5})(2 - \sqrt{5})$.
10. Write $\sqrt[4]{7}$ as a power of $7$, and write $\displaystyle 6^{\frac{1}{3}}$ as a root.

Answers#