Problems on Rationalizing the Denominator

In the previous topics of rational numbers we have learnt to solve the problems regarding the fractional numbers, i.e., the numbers that have real numbers in their denominators. But we haven’t seen much problems regarding those fractions which have irrational numbers in their denominator. Yet I the topic of rationalization we have seen few examples on how to rationalize denominators. Under this topic we’ll see more problems regarding the calculations of rationalization of denominators. Below are given some examples on how to rationalize the complex denominators and proceed further to solve the problems involving these types of complex denominators:-

1. Rationalize $\frac{1}{\sqrt{11}}$ .

Solution:

Since the given fraction has an irrational denominator, so we need to rationalize this and make it more simple. So, to rationalize this, we will multiply the numerator and denominator of the given fraction by root 11, i.e., √11.So,

$\frac{1}{\sqrt{11}}$ $\times$ $\frac{\sqrt{11}}{\sqrt{11}}$

⟹ $\frac{\sqrt{11}}{11}$

So, the required rationalized form of the given denominator is:

$\frac{\sqrt{11}}{11}$ .

2. Rationalize $\frac{1}{\sqrt{21}}$ .

Solution:

The given fraction has an irrational denominator. So, we need to make it simple by rationalizing the given denominator. To do so, we’ll have to multiply and divide the given fraction by root 21, i.e., √21.So,

$\frac{1}{\sqrt{21}}$ $\times$ $\frac{\sqrt{21}}{\sqrt{21}}$

⟹ $\frac{\sqrt{21}}{21}$

So the required rationalized fraction is:

$\frac{\sqrt{21}}{21}$

3. Rationalize $\frac{1}{\sqrt{39}}$ .

Solution:

Since the given fraction has an irrational denominator in it. So, to make the calculations more easy we need to make it simple and hence we need to rationalize the denominator. To do so, we’ll have to multiply both the numerator and denominator of the fraction with root 39, i.e., √39. So,

$\frac{1}{\sqrt{39}}$ $\times$ $\frac{\sqrt{39}}{\sqrt{39}}$

⟹ $\frac{\sqrt{39}}{39}$

So, the required rationalized fraction is:

$\frac{\sqrt{39}}{39}$ .

4. Rationalize $\frac{1}{4+\sqrt{10}}$ .

Solution:

The given fraction consists of irrational denominator. To make the calculations more simplified we will have to rationalize the denominator of the given fraction. To do so, we’ll have to multiply both numerator and denominator by conjugate of the given denominator, i.e., $\frac{4-\sqrt{10}}{4-\sqrt{10}}$ . So,

$\frac{1}{4+\sqrt{10}}$ $\times$ $\frac{4-\sqrt{10}}{4-\sqrt{10}}$

⟹ $\frac{4-\sqrt{10}}{4^{2}-\sqrt{10^{2}}}$

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

⟹ $\frac{4-\sqrt{10}}{16-10}$

⟹ $\frac{4-\sqrt{10}}{6}$

So the required rationalized fraction is:

$\frac{4-\sqrt{10}}{6}$ .

5. Rationalize $\frac{1}{\sqrt{6}-\sqrt{5}}$ .

Solution:

Since the given fraction has irrational denominator in it. So, to make it more simplified we will have to rationalize the denominator of the given fraction. To do so, we’ll have to multiply both numerator and denominator of the fraction by $\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$ . So,

$\frac{1}{\sqrt{6}-\sqrt{5}}$ $\times$ $\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$

⟹ $\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6^{2}}-\sqrt{5^{2}}}$

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

⟹ $\frac{\sqrt{6}+\sqrt{5}}{1}$

⟹ $\sqrt{6}+\sqrt{5}$

So, the required rationalized fraction is:

$\sqrt{6}+\sqrt{5}$

6. Rationalize $\frac{2}{\sqrt{11}-\sqrt{6}}$ .

Solution:

Since, the given fraction has irrational denominator in it which makes the calculations more complex. So, to make them more simplified we’ll have to rationalize the denominator of the given fraction. To do so, we’ll have to multiply both numerator and denominator of the given fraction with $\frac{\sqrt{11}+\sqrt{6}}{\sqrt{11}+\sqrt{6}}$ .