Surds are irrational numbers whose exact values cannot be determined. We can only get the approximate values for surds.

Examples of surds are $\sqrt{17}$, $\sqrt{19}$, $\sqrt{23}$, etc

**Rules Of Operation Of Surds**

The operation of surds are guided by four basic rules

**Rule 1**

If a surd contains two multiples, then we can express each as a different surd. That is,

$$\sqrt{x \times y}=\sqrt{x} \times \sqrt{y}$$

**Rule 2**

If a surd contains a quotient, then we can simplify it differently like this

$$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$$

**Rule 3**

The sum of two numbers in a surd cannot be expressed separately as the sum of two surds.

$$\sqrt{x+y}\neq\sqrt{x}+\sqrt{y}$$

**Rule 4**

The difference of two numbers in a surd cannot be expressed as a difference of two surds

$$\sqrt{x-y}\neq\sqrt{x}-\sqrt{y}$$

**Basic Form Of Surds**

A surd is said to be in **its basic form** if, and if only, it does not have a factor that is a **perfect square**.

For example, $\sqrt{6}$, $\sqrt{7}$, $\sqrt{17}$ are all in their basic forms because they cannot be broken into two or more factors where one is a perfect square

But $\sqrt{80}$ is not in its basic because it has a perfect square in the form of 16. Let express $\sqrt{80}$ in its basic form

$$\sqrt{80}=\sqrt{16\times 5}=4\sqrt{5}$$

Hence $4\sqrt{5}$ is the basic form of $\sqrt{80}$

__Example 1__

Express $\sqrt{40}$ in its basic form

**Solution:**

$\sqrt{40}=\sqrt{4\times 10}=\sqrt{4}\times\sqrt{10}=2\sqrt{10}$

Do you want more examples on expressing surd in basic form? Take a look at this

We can also reverse the process. That is. Write surds from its basic form to its normal form

__Example 2__

Express $2\sqrt{10}$ in its normal form.

**Solution:**

Simply add square root and square to the rational number.

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

**Like Surds**

Like surds are surds that have the same number in their irrational part. They are also called **similar surds.**

For example

$9\sqrt{7}$, $7\sqrt{7}$and $3\sqrt{7}$ are like surds.

Furthermore, $\sqrt{63}$, $\sqrt{28}$ and $\sqrt{252}$ are like surds because each of them can be written as a multiple of $\sqrt{7}$ as follow.

$$\sqrt{63}=\sqrt{9}\times\sqrt{7}=3\sqrt{7}$$

$$\sqrt{28}=\sqrt{4}\times\sqrt{7}=2\sqrt{7}$$

$$\sqrt{252}=\sqrt{36}\times\sqrt{7}=6\sqrt{7}$$

**Conjugate Surds**

Two surds are said to be **conjugate of each other** if their product results in a rational number.

For instance, suppose we multiply

$(\sqrt{a+b})(\sqrt{a-b}=(\sqrt{a})^2-(\sqrt{b})^2$

**Hint**: The conjugate of a surd is simply the additive or subtractive inverse of the surd.

For example,

The conjugate of $\sqrt{7}+\sqrt{5}$ is $\sqrt{7}-\sqrt{5}$

The conjugate of $4\sqrt{6}-3\sqrt{2}$ is $4\sqrt{6}-3\sqrt{2}$

__Example 3__

Simplify $(4-3\sqrt{3})(4+3\sqrt{3})$

**Solution:**

Remember that difference of the two squares states that

$$(a-b)(a+b)=a^2-b^2$$

$$(4-3\sqrt{3})(4+3\sqrt{3})=((4)^2-(3\sqrt{3})^2)$$

$$16-9(3)$$

$$16-27=-11$$

**Addition And Subtraction Of Surds**

Two conditions are necessary to add or subtract surds

1) The surds must be its basic form

2) The surd must be like surds

__Example 4__

Solve $3\sqrt{12}+4\sqrt{108}$

**Solution:**

First, we need to write this expression its basic form

$3\sqrt{4\times 3}+4\sqrt{36\times 3}$

$(3\sqrt{4}\times\sqrt{3})+(4\sqrt{36}\times\sqrt{3})$

$(3\times 2\sqrt{3})+(4 \times 6 \sqrt{3})$

$6\sqrt{3}+24\sqrt{3}$

Since it is now in its basic form and they are like surds, we can now add them

$6\sqrt{3}+24\sqrt{3}=30\sqrt{3}$

__Example 5__

Solve $4\sqrt{99}-\sqrt{44}$

**Solution:**

We begin by expressing it in its most basic form.

$\sqrt{9 \times 11}-\sqrt{4 \times 11}$

$3\sqrt{11}-2 \sqrt{11}=\sqrt{11}$

__Example 6__

Solve $3\sqrt{12}+4\sqrt{72}-5\sqrt{18}+2\sqrt{75}$

**Solution:**

first, we express each in their basic form

$3\sqrt {4 \times 3}+4\sqrt{36\times2}-5\sqrt{9\times 2}+2\sqrt{25\times3}$

$3\times2\sqrt{3}+4\times6\sqrt{2}-5\times 3\sqrt{2}+2\times 5\sqrt{3}$

$6\sqrt{3}+24\sqrt{2}-15\sqrt{2}+10\sqrt{3}$

Now, let add like surds

$6\sqrt{3}+10\sqrt{3}+24\sqrt{2}-15\sqrt{2}$

$16\sqrt{3}+9\sqrt{2}$

As can be observed, we only add like surds

The addition and subtraction of surds have been covered in detail here. So let's move on to the next section.

**Multiplication And Division Of Surds**

As earlier mentioned,

$$\sqrt{x \times y}=\sqrt{x} \times \sqrt{y}$$

$$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$$

The above rule works perfectly for multiplication and division of surds

However, we can extend the preceding rule as follows:

$$a\sqrt{x}\times b\sqrt{y}=(a\times b)(\sqrt{x \times y})=ab\sqrt{xy}$$

$$\frac{a\sqrt{x}}{b\sqrt{y}}=\frac{a}{b}\times\frac{x}{y}$$

__Example 7__

Solve $\sqrt{98}\times\sqrt{96}$

**Solution:**

$\sqrt{98}\times\sqrt{96}=\sqrt{9408}$

However, $\sqrt{9408}$ is not in its basic form

$\sqrt{9408}=\sqrt{3136}\times\sqrt{3}$

$\sqrt{9408}=56\sqrt{3}$

More examples of multiplication of surds can be found here.

__Example 8__

Solve $\sqrt{\frac{7200}{3200}}$

**Solution:**

$\sqrt{\frac{7200}{3200}}=\sqrt{\frac{72}{32}}$

$\sqrt{\frac{72}{32}}=\frac{\sqrt{72}}{\sqrt{32}}$

$\sqrt{\frac{72}{32}}=\frac{\sqrt{36\times 2}}{\sqrt{16\times 2}}$

$\sqrt{\frac{72}{32}}=\frac{6\sqrt{2}}{4\sqrt{2}}$

$\frac{6\sqrt{2}}{4\sqrt{2}}=\frac{3}{2}$

More examples of division of surds can be found here

**Multiplication Of Surds In Bracket**

To multiply surds, we simply expand the surds

__Example 9__

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

**Solution:**

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

$6\sqrt{6}-3\sqrt{4}+2\sqrt{9}-\sqrt{6}$

$6\sqrt{6}-6+6-\sqrt{6}$

$6\sqrt{6}-\sqrt{6}=5\sqrt{6}$

**Rationalization Of Surds**

Rationalization of surd is the process of removing the irrational number from the denominator of the surd.

To achieve this, we multiply the numerator and the denominator of the surd by the conjugate of the denominator

**Related posts**

__Example 10__

Simplify $\frac{5\sqrt{6}-4\sqrt{3}}{3\sqrt{3}+3\sqrt{6}}$

**Solution:**

The conjugate of the denominator is $3\sqrt{3}-3\sqrt{6}$. Hence, we multiply both the denominator and the denominator by $3\sqrt{3}-3\sqrt{6}$

$\frac{5\sqrt{6}-4\sqrt{3}}{3\sqrt{3}+3\sqrt{6}}\times\frac{3\sqrt{3}-3\sqrt{6}}{3\sqrt{3}-3\sqrt{6}}$

$\frac{5\sqrt{6}(3\sqrt{3}-3\sqrt{6})-4\sqrt{3}(3\sqrt{3}-3\sqrt{6})}{(3\sqrt{3})^2-(3\sqrt{6})^2}$

$\frac{15\sqrt{18}-15\sqrt{36}-12\sqrt{9}+12\sqrt{18}}{27-54}$

$\frac{15\sqrt{18}+12\sqrt{18}-90-36}{27-54}$

$\frac{27\sqrt{18}-126}{-27}$

$\frac{81\sqrt{2}-126}{-27}$

$\frac{9(9\sqrt{2}-14)}{-27}$

$\frac{9\sqrt{2}-14}{-3}$

Since it is conventional for the numerator to retain the sign, therefore,

$\frac{-(9\sqrt{2}-140)}{3}$

$\frac{-9\sqrt{2}+140}{3}$

Voila! you've just learnt addition, subtraction, multiplication and division surds.

You have also learnt how to obtain the conjugate of a surd as well as rationalization of surds.

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