In Mathematics, **Surds** are an irrational number which cannot be represented accurately in the form of fractions or recurring decimals. So, it can be left as a square root. Surds are used in many real-time applications to make precise calculations. In this article, let us discuss the surds definition, types, six basic rules of surds, and example problems.

## Surds Definition

Surds are the square rootsÂ (âˆš) of numbers which cannot be simplified into a whole or rational number. It cannot be accurately represented in a fraction. In other words, a surd is a root of the whole number that has an irrational value. Consider an example,Â âˆš2Â â‰ˆ 1.414213. It is more accurate if we leave it as a surdÂ âˆš2.

## Types of Surds

The different types of surds are as follows:

- Simple Surds – A surd that has only one term is called simple surd. Example:Â âˆš2,Â âˆš5, …
- Pure Surds –Â Surds which are completely irrational.Â Example:Â âˆš3
- Similar Surds – The surds having the same common surds factor
- Mixed Surds –Â Surds that are not completely irrational and can be expressed as a product of a rational number and an irrational number
- Compound Surds –Â An expression which is the addition or subtraction of two or more surds
- Binomial Surds –Â Â A surd that is made of two other surds

## Six Rules of Surds

**Rule 1:**

Example:

To simplify âˆš18

18 = 9 x 2 = 3^{2} x 2, since 9 is the greatest perfect square factor of 18.

Therefore, âˆš18 = âˆš(3^{2} x 2)

= âˆš3^{2 }x âˆš2

= 3 âˆš2

**Rule 2:**

Example:

âˆš(12 / 121) = âˆš12 / âˆš121

=âˆš(2^{2} x 3) / 11

=âˆš2^{2} x âˆš3 / 11

= 2âˆš3 / 11

**Rule 3:**

You can rationalize the denominator by multiplying the numerator and denominator by the denominator.

Example:

Rationalise

5/âˆš7

Multiply numerator and denominator by âˆš7

5/âˆš7 = (5/âˆš7) x (âˆš7/âˆš7)

= 5âˆš7/7

**Rule 4:**

Example:

To simplify,

5âˆš6 + 4âˆš6

5âˆš6 + 4âˆš6 = (5 + 4) âˆš6

by the rule

= 9âˆš6

**Rule 5:**

Multiply top and bottom by a-b âˆšn

This rule enables us to rationalise the denominator.

Example:

To Rationalise

\(\frac{3}{2+\sqrt{2}}= \frac{3}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{6-3\sqrt{2}}{4-2}\) \(=\frac{6-3\sqrt{2}}{2}\)**Rule 6:**

This rule enables you to rationalise the denominator.

Multiply top and bottom by a + bâˆšn

Example:

To Rationalise

\(\frac{3}{2-\sqrt{2}}= \frac{3}{2-\sqrt{2}}\times \frac{2+\sqrt{2}}{2+\sqrt{2}}=\frac{6+3\sqrt{2}}{4-2}\) \(=\frac{6+3\sqrt{2}}{2}\)### How to Solve Surds?

You need to follow some rules to solve expressions that involve surds. One method is to rationalize the denominators, which helps to eject the surd in the denominator. Sometimes it may be mandatory to find the greatest perfect square factor to solve surds.

### Surds Problems

**Example 1:**

Write down the conjugate of 5âˆš3 +Â âˆš2

**Solution:**

The conjugate ofÂ 5âˆš3 +Â âˆš2 isÂ 5âˆš3 – âˆš2.

**Example 2:**

Rationalize the denominator: 1/[(8âˆš11 )- (7âˆš5)]

**Solution:**

Given:Â 1/[(8âˆš11 )- (7âˆš5)]

It is known that the conjugate ofÂ (8âˆš11 )- (7âˆš5) isÂ (8âˆš11 )+(7âˆš5)

To rationalize the denominator of the given fraction, multiply the conjugate of denominator on both numerator and denominator.

=[1/[(8âˆš11 )- (7âˆš5)]]Ã— [[(8âˆš11 )+ (7âˆš5)]/[(8âˆš11 )+(7âˆš5)]]

=[(8âˆš11 )+ (7âˆš5)]/[(8âˆš11 )^{2}-(7âˆš5)^{2}]

=[(8âˆš11 )+ (7âˆš5)]/[704- 245]

=Â [(8âˆš11 )+ (7âˆš5)]/459

Now to start practising problems and examples of surd based on rules mentioned above, please visit BYJU’S – The Learning App.

is this surd or not : sqrt ( 3+ (sqrt 2) ) . Give proper reason.

sqrt ( 3+ (sqrt 2) ) is a surd because it cannot be simplified into a whole or rational number.

Hi mj

It is a surd

cuz it is an irrational number

This is a surd as it is an irrational number. Rational numbers have either the terminating decimal places or non terminating but repeating decimal places. sqrt ( 3+ (sqrt 2) ) the decimal places of the answer never terminates and never repeats, thus it is an irrational number and a surd

Please help me solve this question

5âˆš6+3âˆš24-6âˆš54+âˆš150

5âˆš6+3âˆš24-6âˆš54+âˆš150

= 5âˆš6+3.4âˆš6-6.3âˆš6+5âˆš6

= 5âˆš6+12âˆš6-18âˆš6+5âˆš6

= (5+12-6+5)âˆš6

= 16âˆš6

5âˆš6+3âˆš24-6âˆš54+âˆš150

=5âˆš6+3âˆš4*6-6âˆš9*6+âˆš25*6

=5âˆš6+3*2âˆš6-6*3âˆš6+5âˆš6

=5âˆš6+6âˆš6-18âˆš6+5âˆš6

=-2âˆš6

5root6+6root6-18root6+5root6

16root6-18root6

=-2root6