Simple and Compound Surds

We will discuss about the simple and compound surds.

Definition of Simple Surd:

A surd having a single term only is called a monomial or simple surd. 

Surds which contains only a single term, are called as nominal or simple surds. For example $\sqrt[2]{2}$, $\sqrt[2]{5}$,$\sqrt[2]{7}$, $5\sqrt[3]{10}$, $3\sqrt[4]{12}$, $a\sqrt[n]{x}$ are simple surds.

More example, each of the surds √2, ∛7, ∜6, 7√3, 2√a, 5∛3, m∛n, 5 ∙ 7$^{3/5}$ etc. is a simple surd.

Definition of Compound Surd:

The algebraic sum of two or more simple surds or the algebraic sum of a rational number and simple surds is called a compound scud. 

The algebraic sum of two or more simple surds or the algebraic sum of rational numbers and simple surds are called as binominal surds or compound surds. For example $2+\sqrt[2]{3}$ is a sum of one rational number 2 and one simple surd $\sqrt[2]{3}$, so this is a compound surd. $\sqrt[2]{2} + \sqrt[2]{3}$ is a sum of two simple surds $\sqrt[2]{2}$ and $\sqrt[2]{3}$, so this is also a example of compound surd. Some other examples of compound surds are $\sqrt[2]{5} -\sqrt[2]{7}$, $\sqrt[3]{10} + \sqrt[3]{12}$, $\sqrt[2]{x} + \sqrt[2]{y}$

More example, each of the surds (√5 + √7), (√5 - √7), (5√8 - ∛7), (∜6 + 9), (∛7 + ∜6), (x∛y - b)  is a compound surd. 

Note: The compound surd is also known as binomial surd. That is, the algebraic sum of two surds or a surd and a rational number is called a binomial surd. 

For example, each of the surds (√5 + 2), (5 - ∜6), (√2 + ∛7) etc. is a binomial surd.

Problems on Simple Surds:

1. Arrange the following simple surds descending order.

$\sqrt[2]{3}$, $\sqrt[3]{9}$,$\sqrt[4]{60}$

Solution:

The given surds are $\sqrt[2]{3}$, $\sqrt[3]{5}$, $\sqrt[4]{12}$.

The surds are in the order of 2, 3, and 4 respectively. If we need to compare their values, we need to express them in same order. As the LCM of 2, 3, and 4 is 12, we should express the surds in order 12.

$\sqrt[2]{3}$ = $3^{\frac{1}{2}}$ = $3^{\frac{6}{12}}$= $729^{\frac{1}{12}}$ = $\sqrt[12]{729}$

$\sqrt[3]{5}$ = $5^{\frac{1}{3}}$ = $5^{\frac{4}{12}}$= $625^{\frac{1}{12}}$ = $\sqrt[12]{625}$

$\sqrt[4]{12}$ = $12^{\frac{1}{4}}$ = $12^{\frac{3}{12}}$ = $1728^{\frac{1}{12}}$ = $\sqrt[12]{1728}$

Hence the descending order of the given surds is $\sqrt[4]{12}$, $\sqrt[2]{3}$, $\sqrt[3]{5}$.

2. Arrange the following simple surds descending order.

$2\sqrt[2]{10}$, $4\sqrt[2]{7}$, $5\sqrt[2]{3}$

Solution:

If we need to compare the values of the given simple surds, we have to express them in the form of pure surds. As the orders of all three surds are same we don’t need change the order.

$2\sqrt[2]{10}$ = $\sqrt[2]{2^{2}\times 10}$ = $\sqrt[2]{4\times 10}$ = $\sqrt[2]{40}$

$4\sqrt[2]{7}$ = $\sqrt[2]{4^{2}\times 7}$ = $\sqrt[2]{16\times 7}$ = $\sqrt[2]{112}$

$5\sqrt[2]{3}$ = $\sqrt[2]{5^{2}\times 3}$ = $\sqrt[2]{25\times 3}$= $\sqrt[2]{75}$

Hence the descending order of the given surds is $4\sqrt[2]{7}$, $5\sqrt[2]{3}$, $2\sqrt[2]{10}$.

Problems on Compound Surds:

1. If x = $1+\sqrt[2]{2}$, then what is the value of $x^{2} - \frac{1}{x^{2}}$?

Solution:

Given x = $1+\sqrt[2]{2}$

We need find out 

$x^{2}-\frac{1}{x^{2}}$

= $x^{2}-(\frac{1}{x})^{2}$

As we know $a^{2}-b^{2} = (a + b)(a - b)$

We can write $x^{2} - (\frac{1}{x})^{2}$ as

= $(x + \frac{1}{x})(x - \frac{1}{x})$

Now we will find out separately the values of  $x+\frac{1}{x}$ and  $x-\frac{1}{x}$

$x+\frac{1}{x}$

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

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

=$\frac{1+2+2\sqrt{2}+1}{1+\sqrt{2}}$

=$\frac{4+2\sqrt{2}}{1+\sqrt{2}}$

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

=$2\sqrt{2}$$x-\frac{1}{x}$

=$1+\sqrt[2]{2}$-$\frac{1}{1+\sqrt{2}}$

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

=$\frac{1+2+2\sqrt{2}-1}{1+\sqrt{2}}$

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

So $x^{2} - \frac{1}{x^{2}}$

=$(x+\frac{1}{x})\cdot (x-\frac{1}{x})$

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

=$\frac{6\sqrt{3}+8}{1+\sqrt{2}}$

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

2. If x= $\sqrt{2}+\sqrt{3}$ and y = $\sqrt{2}-\sqrt{3}$ then what is the value of $x^{2}- y^{2}$?

Solution:

As we know $a^{2}-b^{2} = (a+ b)(a - b)$

$x^{2}- y^{2}$

= $(x+y)(x-y)$

Now we will find out separately the values of (x + y) and (x - y).

   (x + y)

= $\sqrt{2}+\sqrt{3}$ + $\sqrt{2}-\sqrt{3}$

= $2\sqrt{2}$(x - y)

= $\sqrt{2} + \sqrt{3}$-$\sqrt{2} - \sqrt{3}$

= $2\sqrt{3}$

So $x^{2}- y^{2}$

= $2\sqrt{2}\times2\sqrt{3}$

=$4\sqrt{6}$

11 and 12 Grade Math

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