Question

Rationalisation of denominator

$\sf \large \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} }$
​

Answer 1

Solution :

We have -

$\begin{aligned} \dfrac{7 - 3 \sqrt 5}{3 - \sqrt 5} &= \dfrac{(7 - 3 \sqrt 5)}{(3 - \sqrt 5)} \times \dfrac{(3 + \sqrt 5)}{(3 + \sqrt 5)} = \dfrac{(7 - 3 \sqrt 5)(3 + \sqrt 5)}{3^2 - (\sqrt 5)^2}\\\\\\&\Rightarrow \dfrac{21 + 7 \sqrt 5 - 9 \sqrt 5 - 15}{9 - 5} = \dfrac{6 - 2 \sqrt 5}{4}\\\\\\&\Rightarrow \dfrac{2 (3 - \sqrt 5)}{4}= \bf \dfrac{3 - \sqrt 5}{2} \end{aligned}$

$\begin{gathered} \end{gathered}$

Here, the Rationalising Factor is $\mathrm{\bf 3 + \sqrt 5}$

$\begin{gathered} \end{gathered}$Hence, the Rationalized Form of $\dfrac{7 - 3 \sqrt 5}{3 - \sqrt 5}$ is $\bf \dfrac{3 - \sqrt 5}{2}$ . . .

$\begin{gathered} \end{gathered}$

Additional Information :

Rationalization :-

It is the Process of converting a Number whose Denominator is Irrational, into an Equivalent Expression with a Rational Denominator.

$\begin{gathered} \end{gathered}$

This is done by Multiplying the Number [i.e., Both Numerator and the Denominator] by a Number which is called as Rationalising Factor.

Answer 2

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