Question

Kindly rationalise this:-
$\boxed{\huge{\sf{ \dfrac{x}{2 \sqrt{x - 1} } }}}$
​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Rationalization has been used. There's a common doubt among everyone that what does the term rationalization mean. This means simplifying a given term into its simplest form. It can include many algebraic identities or multiplying the numerator and denominator by same term to get the simplest form. Using this concept, we can find our answer.

Let's do it !!

________________________________________________

★ Solution :-

$\\\;\bf{\mapsto\;\;\green{\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}}}$

We shall do this question step by step.

• Step 1.) ::

We see that in the given fraction, the denominator is a product of two terms. So firstly in order to make this rationalization easier, we shall seperate the terms in such a way that its value doesn't change.

The given expression can we written as,

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\blue{\dfrac{x}{2}\;\times\;\dfrac{1}{\:\sqrt{x\:-\:1}\:}}}}$

We see that the value of given equation doesn't change.

• Step 2.) ::

For rationalization, firstly we have to remove the (√) square root from the denominator.

So, in this step we shall multiply the numerator and denominator by a common term which makes the denominator to its square.

So, this can be written as

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\dfrac{x}{2}\;\times\;\dfrac{1}{\:\sqrt{x\:-\:1}\:}\;\times\;\dfrac{\sqrt{x\:-\:1}}{\sqrt{x\:-\:1}}}}$

Multiplying both numerator and denominator by same term won't change the value of the given expression.

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\pink{\dfrac{x}{2}\;\times\;\dfrac{\sqrt{x\:-\:1}}{\:(\sqrt{x\:-\:1})(\sqrt{x\:-\:1})}}}}$

• Step 3.) ::

Now we have headed towards rationalization. Further the simplification goes one like this,

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\dfrac{x}{2}\;\times\;\dfrac{\sqrt{x\:-\:1}}{\:(\sqrt{x\:-\:1})^{2}}}}$

(since two equal terms are being multiplied)

Now we can remove the square root because the terms are being squared.

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\red{\dfrac{x}{2}\;\times\;\dfrac{\sqrt{x\:-\:1}}{(x\:-\:1)}}}}$

• Step 4.) ::

This is the almost final step. Here we shall combine the terms to get final answer.

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\dfrac{x\:\sqrt{x\:-\:1}}{2(x\:-\:1)}}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\orange{\dfrac{x\:\sqrt{x\:-\:1}}{2x\:-\:2}}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Required\;\:rationalisation\;=\;\bf{\purple{\dfrac{x\:\sqrt{x\:-\:1}}{2x\:-\:2}}}}}}$

________________________________________________

★ More to know :-

$\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:2ab}$

$\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}$

$\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}$

$\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}$

$\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}$

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Rationalization has been used. There's a common doubt among everyone that what does the term rationalization mean. This means simplifying a given term into its simplest form. It can include many algebraic identities or multiplying the numerator and denominator by same term to get the simplest form. Using this concept, we can find our answer.

Let's do it !!

________________________________________________

★ Solution :-

$\\\;\bf{\mapsto\;\;\green{\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}}}$

We shall do this question step by step.

• Step 1.) ::

We see that in the given fraction, the denominator is a product of two terms. So firstly in order to make this rationalization easier, we shall seperate the terms in such a way that its value doesn't change.

The given expression can we written as,

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\blue{\dfrac{x}{2}\;\times\;\dfrac{1}{\:\sqrt{x\:-\:1}\:}}}}$

We see that the value of given equation doesn't change.

• Step 2.) ::

For rationalization, firstly we have to remove the (√) square root from the denominator.

So, in this step we shall multiply the numerator and denominator by a common term which makes the denominator to its square.

So, this can be written as

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\dfrac{x}{2}\;\times\;\dfrac{1}{\:\sqrt{x\:-\:1}\:}\;\times\;\dfrac{\sqrt{x\:-\:1}}{\sqrt{x\:-\:1}}}}$

Multiplying both numerator and denominator by same term won't change the value of the given expression.

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\pink{\dfrac{x}{2}\;\times\;\dfrac{\sqrt{x\:-\:1}}{\:(\sqrt{x\:-\:1})(\sqrt{x\:-\:1})}}}}$

• Step 3.) ::

Now we have headed towards rationalization. Further the simplification goes one like this,

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\dfrac{x}{2}\;\times\;\dfrac{\sqrt{x\:-\:1}}{\:(\sqrt{x\:-\:1})^{2}}}}$

(since two equal terms are being multiplied)

Now we can remove the square root because the terms are being squared.

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\red{\dfrac{x}{2}\;\times\;\dfrac{\sqrt{x\:-\:1}}{(x\:-\:1)}}}}$

• Step 4.) ::

This is the almost final step. Here we shall combine the terms to get final answer.

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\dfrac{x\:\sqrt{x\:-\:1}}{2(x\:-\:1)}}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{\:x\:}{\:2\sqrt{x\;-\;1}\:}\;=\;\bf{\orange{\dfrac{x\:\sqrt{x\:-\:1}}{2x\:-\:2}}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Required\;\:rationalisation\;=\;\bf{\purple{\dfrac{x\:\sqrt{x\:-\:1}}{2x\:-\:2}}}}}}$

________________________________________________

★ More to know :-

$\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:2ab}$

$\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}$

$\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}$

$\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}$

$\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}$