Simple Surds and Compound Surds

The simple surds and the compound surds are two different types of surds. Here we will learn about them.

Definition of a Simple Surd:

A surd is called a simple surd if it contains only one term. More precisely, this type of surd is produced with only one root symbol. Simple surd is also known as a monomial surd.

Note that $5^{3/2}$ $=\sqrt{5^3}$$=\sqrt{125}$ contains only one term (or only one root symbol). So by the definition, $5^{3/2}$ is a simple surd or a monomial surd.

Here we list more examples of simple surds.

 

Examples of Simple Surds:

(i) $\sqrt{2},$ $\sqrt{3},$ $\sqrt{7}$ are simple surds or monomial surds as each of them consists of only one term.

(ii) Note that $2.7^{2/3}$ $=2.\sqrt[3]{7^2}$ $=2.\sqrt[3]{49}$ $=\sqrt[3]{2^3 \times 49}.$ Thus $2.7^{2/3}$ contains only one real root. Hence it is an example of simple surds.

(iii) $\sqrt[5]{7},$ $7\sqrt{5},$ $\sqrt[n]{a}$ all are examples of simple surds or monomial surds.

(iv) Similarly, $5\sqrt[5]{16},$ $32^{5/2},$ $27^{3/2},$ $8^{-3/2}$ are simple surds or monomial surds.

 

Next, we will discuss about the compound surds.

Definition of a Compound Surd:

A surd is called a compound surd if it is the algebraic sum or difference of either of the following two:

(i) two or more simple surds

(ii) rational numbers and simple surds.

 

Remarks on Compound Surds:

• The sum or the difference of two simple surds is called a compound surd.

• The sum or the difference of a rational number and simple surds is also said to be a compound surd.

• The compound surds are also often called the binomial surds when they are produced with two terms only. So the binomial surd is a combination of either of (i) two simple surds or (ii) a rational number and a simple surd.

 

Examples of Compound Surds:

(i) $1+\sqrt{5}$ is a sum of a rational number $1$ and a simple surd $\sqrt{5}.$ So $1+\sqrt{5}$ is a compound surd. Again since $1+\sqrt{5}$ is a combination of two terms, it is a binomial surd.

(ii) Note that $2-\sqrt{3}+\sqrt{5}$ is a combination of one rational number and two simple surds. Thus it is a compound surd.

(iii) $\sqrt{2}+\sqrt{3}+\sqrt{7}$ is a sum of three simple surds $\sqrt{2},$ $\sqrt{3}$ and $\sqrt{7}.$ So it is a compound surd.

(iv) Similarly, $\sqrt{2}-\sqrt{7},$ $3\sqrt{2}+2\sqrt{3},$ $\sqrt[3]{3} \pm \sqrt[5]{10},$ $\sqrt[n]{a} \pm \sqrt[m]{b}$ all are examples of compound surds or binomial surds.

 

Solved Problems on Simple and Compound Surds:

Problem 1: Determine the type of the following:

(i) $4^{3/5}$

(ii) $2 . 4^{3/5}$

(iii) $2+4^{3/5}$

(iv) $\frac{1}{5}$

(v) $\sqrt{1+\sqrt{5}}$

Solution:

(i) Note that $4^{3/5}$ $=(4^3)^{1/5}$

$=\sqrt[5]{4^3}$

$=\sqrt[5]{64}$

So it is a simple surd.

(ii) Again $2 . 4^{3/5}$

$=2. \sqrt[5]{64}$ [by part (i)]

$=\sqrt[5]{2^5\times 64}$

Thus $2.4^{3/5}$ contains only one term, and so it is a simple surd.

(iii) By part (i), we get that $4^{3/5}$ is a simple surd.

So $2+4^{3/5}$ is a sum of the rational number $2$ and the simple surd $4^{3/5}$. As a result, we can say that it is a compound surd.

(iv) Multiplying both the numerator and the denominator of $\frac{1}{\sqrt{5}}$ by $\sqrt{5},$ we get that

$\frac{1}{\sqrt{5}}$ $=\frac{1 \times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$ $=\frac{1}{5}\sqrt{5}$

So it contains only one root symbol, and it is a simple surd.

(v) Note that the sum of a rational number and an irrational number is an irrational number. Now, as $\sqrt{5}$ is an irrational number we must have that $1+\sqrt{5}$ is an irrational number. Thus the given number is the square root of an irrational number. So we conclude that it is NOT a surd.

 

Problem 2: Show that $\frac{1}{\sqrt{5}-1}$ is a compound surd.

Solution:

To show the given surd is a compound surd, we need to first rationalize the denominator. We multiply both the numerator and the denominator of $\frac{1}{\sqrt{5}-1}$ by $\sqrt{5}-1.$ Thus we obtain that

$\frac{1}{\sqrt{5}-1}$ $=\frac{1 \times (\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}$

$=\frac{\sqrt{5}+1}{5-1}$ $[\because (a-b)(a+b)=a^2-b^2]$

$=\frac{1}{4}(\sqrt{5}+1)$

As $\sqrt{5}+1$ is a compound surd, we deduce that the given surd is a compound surd.

 

Problem 3: Show that the reciprocal of a simple surd is again a simple surd.

Solution:

Note that the general form of a simple surd is $a\sqrt[n]{b},$ where $n$ is an integer and both $a, b$ are rational numbers. Now the reciprocal of $a\sqrt[n]{b}$ is

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

$=\frac{1}{ab^{1/n}}$

$=\frac{1}{a}b^{-1/n}$

$=\frac{1}{a}\sqrt[-n]{b}$

Thus it contains only one root symbol. So the reciprocal of a simple surd is again a simple surd.

 

Problem 4: Find the square root of the compound surd $2+\sqrt{3}$

Solution: Note that

$2+\sqrt{3}$

$=\frac{1}{2}[2(2+\sqrt{3})]$

$=\frac{1}{2}(4+2\sqrt{3})$

$=\frac{1}{2}(1+3+2\sqrt{3})$

$=\frac{1}{2}(1^2+\sqrt{3}^2+2. 1 .\sqrt{3})$

$=\frac{1}{2}(1+\sqrt{3})^2$ $[\because a^2+b^2+2ab=(a+b)^2]$

So the square root of $2+\sqrt{3}$ is $\pm\frac{1}{\sqrt{2}}(1+\sqrt{3}).$

 

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