Question

i can't understand the concept of rationalisation of class 9th . can anyone describe with examples​

Answer 1

Answer :-

When we come across expression having square roots in their denominator. To make the denominator free from that square roots we multiply the numerator and denominator by an irrational number.

Let's take a example

$\frac{2}{ \sqrt{3} }$

⇒

$\frac{2}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{2 \sqrt{3} }{3}$

Another example

⇒

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

$\frac{1}{3 + \sqrt{2} } \times \frac{3 - \sqrt{2} }{3 - \sqrt{2} } = \frac{3 - \sqrt{2} }{9 - 2} = \frac{3 - \sqrt{2} }{7}$

Answer 2

Rationalisation

Suppose, in a problem you come across a number like $\sf \frac{1}{\sqrt{2}}$. Now, you don't like seeing the square root of 2 in denominator. You are willing to see that √2 in numerator. How can we do that?

⇒ Rationalisation is the process by which we change the irrational denominator of a fraction, into rational denominator.

Now, let's try changing $\sf\frac{1}{\sqrt{2}}$ in a fraction having rational denominator.

If we will multiply √2 on both numerator and denominator then :-

$\sf\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

Now, since we are multiplying by $\frac{\sqrt{2}}{\sqrt{2}}$, it won't affect value because $\frac{\sqrt{2}}{\sqrt{2}}$  = 1.

Continuing, we get

$\sf\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\ \\ \implies \frac{\sqrt{2}}{2}$

Since, √2 ×√2 =2, and 2 is a rational number. So we got a rational denominator without changing the actual value! This is called as rationalisation, and the number we used to change it into rational, is called as rationalising factor. Here, √2 was rationalising factor.

What if there are two terms in denominator?

Let's try to solve them!

$\sf\frac{1}{\sqrt{6}-\sqrt{5}}$

Lo! Now if we will multiply it by √6 - √5, then it would become (√6 - √5)² in denominator and when we will solve it, it will become 6 + 5 - 2√6×√5 = 11 - 2√30. And again, 2√30 is irrational! So what should we do?

What about multiplying with √6 + √5? Would it become (√6 - √5)(√6 + √5)? And wait, this seems to be known! Yes, it's of the form of (a - b)(a + b), which gives a² - b². So, (√6 - √5)(√6 + √5) = (√6)² - (√5)² = 6 - 5 = 1.

Voila! This is what we wanted, a rational number and 1 is a rational number.

So, let's solve it!

$\sf\frac{1}{\sqrt{6}-\sqrt{5}}\\ \\ \implies\frac{(\sqrt{6} + \sqrt{5})}{(\sqrt{6}-\sqrt{5})(\sqrt{6} + \sqrt{5})}\\ \\ \implies \frac{\sqrt{6} + \sqrt{5}}{6 - 5}\\ \\ \implies \frac{\sqrt{6} + \sqrt{5}}{1} = \sqrt{6} + \sqrt{5}$

So, we conclude that if the denominator has only one term, we would multiply by the same for rationalisation, but if it has two terms, we would multiply it by changing the sign between them. That is,

• √a can be rationalised by multiplying with √a (rationalising factor)

• √a + √b can be rationalised by √a - √b

• √a - √b can be rationalised by √a + √b

• a + √b can be rationalised by a - √b

• a - √b can be rationalised by a + √b