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#### Rationalization of Surds

If the product of two surds is rational, then each of them is called the rationalizing factor of the other. The product of √3+√2 and √3-√2 is 3 which is a rationalizing factor of √3-√2 and vice-versa. √3+√2 and√3-√2 are conjugate to each other. So, for a binomial quadratic surd, its conjugate is the rationalizing factor. Note that √5 is rationalized when it is multiplied by √5 so √5 is rationalizing factor of √5.

Since 2√5 x√5 = 10, then
2√5 and √5 are rationalizing factor of each other.
Let us took at another example.
2√5 x 3√5 =30
∴ 2√5 and 3√5 are rational factors of each other. Thus, surd may have more than one rational factor. But it is better to choose a rational factor that is easy to evaluate.

#### Equation involving Surds

The equations are which the variables are expressed in terms of surds are called the equations involving surds.$$\sqrt{3x+2}$$ +$$\sqrt{3x-11}$$ =9 is an example of equation involving surds.

The steps of solving surd equations are as follows:

1. Keep one term of the surds equation under the radical sign on one side and rest on the other side.
2. Raise both sides to the index equal to the degree of surds.
3. Continue the process till the variable comes free from the radical sign.

In the solution of equation involving radical sign, verification must be shown. If any value of the variable doesn't satisfy the given equation, it should be discarded.

• If the product of two surds is rational, then each of them is called the rationalizing factor of the other.
• The Surd may have more than one rational factor.
• The equations are which the variables are expressed in terms of surds are called the equations involving surds.
• If any value of the variable doesn't satisfy the given equation, it should be discarded.
.

#### Click on the questions below to reveal the answers

Solution:

$$\sqrt{ x}$$-2 = 5

or, $$\sqrt{x}$$ = 3+2

or, $$\sqrt{x}$$ = 5

Squaring on both sides,

or, $$\sqrt{x}$$2 =(5)2

or, x = 25

Checking for x = 25,

$$\sqrt{25}$$ - 2 = 3

or, 5 - 2 = 3

or, 3 = 3 (true)

Hence, x = 25.

Solution:

Here, $$\frac{3+ \sqrt{6}}{\sqrt{3 + \sqrt{2}}}$$  =a + b $$\sqrt{3}$$

or, $$\frac{3+ \sqrt{6}}{\sqrt{3 + \sqrt{2}}}$$ $$\times$$ $$\frac{\sqrt3- \sqrt{2}}{\sqrt{3 - \sqrt{2}}}$$= a+b$$\sqrt{3}$$

or, $$\frac{3 \sqrt{3} - 3\sqrt{2} + \sqrt{18} - \sqrt{12}} {3-2}$$ = a + b $$\sqrt{3}$$

or, 3$$\sqrt{3} - 3 \sqrt{2} + 3\sqrt{2} - 2 \sqrt{3}$$ = a + b $$\sqrt{3}$$

or, $$\sqrt{3}$$ = a + b $$\sqrt{3}$$

or, 0 + 1. $$\sqrt{3}$$ = a + b $$\sqrt{3}$$

Comparing the coefficients, we get a =0 and b=1

∴ a = 0 and b= 1

Solution:

$$\frac{\sqrt{a + 1} - \sqrt{a^2 -1}} {\sqrt{a^2 + 1} + \sqrt{a^2 -1}}$$ $$\times$$ $$\frac{\sqrt{a + 1} - \sqrt{a^2 -1}} {\sqrt{a^2 + 1} - \sqrt{a^2 -1}}$$

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

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

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

=$$\frac{2a^2-2\sqrt{{a^4}-{1}}}{2}$$

= $$\frac{2( a^2 - \sqrt{a^4 -1)}}{2}$$

= $$a^2 - \sqrt{a^4 - 1}$$

Solution:

$$\frac{1}{\sqrt{7} -1}$$

= $$\frac{1}{\sqrt{7} -1}$$ $$\times$$ $$\frac{\sqrt{7} +1}{\sqrt{7}+1}$$

= $$\frac{\sqrt{7} +1}{(\sqrt{7})-(1)^2}$$

= $$\frac{\sqrt{7} +1}{7-1}$$

=  $$\frac{\sqrt{7} +1}{6}$$

0%

x = 12
x = 15
x = 11
x = 13
• ### Find the square root of 13 + 2 (sqrt{30})

(sqrt{10}) + (sqrt{7})
(sqrt{10}) + (sqrt{3})
(sqrt{15}) + (sqrt{2})
(sqrt{10}) + (sqrt{5})

x = 7
x= 9
x = 5
x=11

(sqrt{2}) +1
(sqrt{2}) +2
(sqrt{3}) +2
(sqrt{3}) +1

x = 1 and 10
x = 1 and 12
x = 1 and 8
x = 2 and 7

x =4
x=5
x= 7
x= 3

x = 30
x = 20
x = 25
x = 15
• ### What is the rationalizing factor of (sqrt{x + a})

(sqrt{1+ a})
(sqrt{x - a})
(sqrt{x + a})
(sqrt{1 - a})

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