Surds: Definition, Rules, Types, and Solved Examples

In this section, we will discuss what are surds with their properties, solved problems, types of surds, and many more.

Definition of a Surd

A root of a positive real number is called a surd if we cannot remove the root symbol after simplification.

Examples of surds: Note that we cannot remove the root symbol from $\sqrt{2}, \sqrt{3}$; so by definition they are surds. But $\sqrt{9}$ is not a surd as its value is $3.$

 

Some Remarks about Surds

  From the above example $\sqrt{9}$, we see that all roots may not be surds.

  All surds are irrational numbers, but the converse is not true. We know that both $\pi$ and $e$ are irrational numbers; but they are not surds.

 

Order of Surds

The order of a surd is the index of the root/radical. For example, the order of $\sqrt[4]{2}$ is $2$.

  A surd of order $2$ is called a quadratic surd.  For example, $\sqrt{2}$ is a quadratic surd.

  A surd of order $3$ is called a cubic surd. For example, $\sqrt[3]{2}$ is a cubic surd.

Ex 1: Express $\sqrt{2}$ as a surd of order $4$.

Solution: $\sqrt{2}$ $=2^{\frac{1}{2}}$ $=2^{\frac{1 \times 2}{2 \times 2}}$ $=2^{\frac{2}{4}}$ $=(2^2)^{\frac{1}{4}}$ $=4^{\frac{1}{4}}$ $=\sqrt[4]{4}$

So we can express $\sqrt{2}$ as $\sqrt[4]{4}$ which is a surd of order $4.$

 

Rules of Surds

Let $a$ and $b$ are two positive real numbers. The following rules are satisfied by the surds.

  $\sqrt{a}=a^{\frac{1}{2}}$

  $\sqrt[n]{a}=a^{\frac{1}{n}}$

  Square root of a square: $\sqrt{a^2}=a$

  Product of two equal square roots: $\sqrt{a} \times \sqrt{a}=a$

  Product of two square roots: $\sqrt{a} \times \sqrt{b}=\sqrt{a \times b}$

  Division of two square roots: $\sqrt{a} \div \sqrt{b}=\sqrt{\frac{a}{b}}$

 

Types of Surds

There are various types of surds. Here we list all of them with examples.

  Simple Surd: This type of surd contains only one term. For example, $\sqrt{5}$ is a simple surd.

  Compound Surd: The sum of two or more simple surds is an example of a compound surd. So $\sqrt{5}+\sqrt{7}$ is a compound surd.

  Pure Surd: If a whole rational number is under $\sqrt{}$, then it is a pure surd. For example, $\sqrt{6}$ is a pure surd. But $3\sqrt{2}$ is not a pure surd as $3$ is not under the root symbol. Note that $3\sqrt{2}$ is a mixed surd.

  Mixed Surd: It’s a product of a rational number and a surd. For example, $2\sqrt{7}$ is a mixed surd.

  Similar Surds: Two surds are said to be similar if their surds factors are common. For example, $\sqrt{2}$ and $5\sqrt{2}$ are similar surds as the only surd factor $\sqrt{2}$ is common in both of them. Also, $\sqrt{2}$ and $\sqrt{18}$ are similar as $\sqrt{18}$ $=3\sqrt{2}.$

  Dissimilar Surd: Two surds are said to be dissimilar if they have different surd factors. For example, $3\sqrt{2}$ and $5\sqrt{3}$ are dissimilar surds.

  Binomial Surd: This type of surd is produced with the help of two surds.

 

Solved Problems of surds

Problem 1: Determine whether $\sqrt{2} \times \sqrt{3}$ is a surd or not.

Solution. Note that $\sqrt{2} \times \sqrt{3}$ $=\sqrt{2 \times 3}$ $=\sqrt{6}$

So $\sqrt{2} \times \sqrt{3}$ is a surd.

 

Problem 2: Find the square root of $125.$

Solution. Note that $125$ $=5 \times 5 \times 5$

$\therefore \sqrt{125}$ $=\sqrt{5 \times 5 \times 5}$

$=\sqrt{5 \times 5} \times \sqrt{5}$ $[\because \sqrt{a \times b}=\sqrt{a} \times \sqrt{b}]$

$=5 \times \sqrt{5}$ $[\because \sqrt{a \times a}=a]$

$=5\sqrt{5}$

 

Problem 3: Find the cube root of $125.$

Solution. As $125$ $=5 \times 5 \times 5=5^3$, we have

$\sqrt[3]{125}$ $=\sqrt[3]{5^3}$

$=(5^3)^{1/3}$ $[\because \sqrt[3]{a}=a^{1/3}]$

$=5^{3 \times \frac{1}{3}}$ $[\because (a^m)^n=a^{m \times n}]$

$=5^1=5$

So the cube root of $125$ is $5.$

 

Problem 4: Rationalize the denominator of $\frac{1}{\sqrt{7}}$

Solution. Multiplying the numerator and the denominator of $\frac{1}{\sqrt{7}}$ by $\sqrt{7},$ we get

$\dfrac{1}{\sqrt{7}}$ $=\dfrac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}$

$=\dfrac{\sqrt{7}}{7}$ $[\because \sqrt{a} \times \sqrt{a}=a]$

 

Problem 5: Rationalize the denominator of $\dfrac{1}{1+\sqrt{2}}$

Solution. Multiplying the numerator and the denominator of $\frac{1}{1+\sqrt{2}}$ by $1-\sqrt{2},$ we get

$\dfrac{1}{1+\sqrt{2}}$

$=\dfrac{1 \times (1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}$

$=\dfrac{1-\sqrt{2}}{1^2-(\sqrt{2})^2}$  $[\because (a+b)(a-b)=a^2-b^2]$

$=\dfrac{1-\sqrt{2}}{1-2}$

$=-1+\sqrt{2}$

 

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