Question

Rationalize the denominator

Answer 1

Let's see the procedure of rationalisation.

To understand the concept clearly, let's take an irrational number to rationalise.

Let the irrational fraction be $\sf \frac{x + \sqrt{y}}{x - \sqrt{y}}$

Now, to rationalise the denominator, multiply both sides with the denominator with positive symbol, i.e., x + √y.

We have :

$\frac{x + \sqrt{y} }{x - \sqrt{y} } \times \frac{x + \sqrt{y} }{x + \sqrt{y} } \\ \\ \frac{(x + \sqrt{y}) {}^{2} }{(x) {}^{2} - ( \sqrt{y} ){}^{2}} \\ \\ \frac{x {}^{2} + y + 2x \sqrt{y} }{x {}^{2} - y }$

Here, after rationalising the denominator, we get x² - y as a denominator, which is an irrational number. Hence, we are done.

Identities used here :

( a + b )² = a² + b² + 2ab

( a + b ) ( a - b ) = a² - b²

Answer 2

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Here is your answer
If we want to rationalize the denominator of any numbers we have to multiply it by it's conjegate on both sides

EXAMPLE
IF THEIR IS IN DENOMINATOR NUMBER
$1. \: \: (x + \sqrt{y}) \: we \: mutiply \: it \: by \: \\ (x - \sqrt{y}) \\ 2. \: \: \: ( \sqrt{x} - \sqrt{y} ) \: \: than \: we \: multiply \: \\ by \: (( \sqrt{x} + \sqrt{y} ) \\ 3. \: \: \: ( \sqrt{x} + \sqrt{y} ) \: \: than \: we \: multiply \: it \: \\ by \: ( \sqrt{x} - \sqrt{y} ) \\ 4. \: \: (x - \sqrt{y} ) \: \: than \: w \: multiply \: it \: by \\ (x + \sqrt{y} )$

SIMILARLY WE RATIONALIZE THE ANY DENOMINATOR.

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