Question

4+3√5÷4-3√5 = a+b √5

Answer 1

Question :–

$\\ \: { \huge{.}} \: \: { \bold{if \: \: \: \dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \: \: = a + b \sqrt{5} \: \: then \: \: find \: \: a \: \: and \: \: b}}$

ANSWER :–

GIVEN :–

$\\ \: { \huge{.}} \: \: { \bold{\: \: \dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \: \: = a + b \sqrt{5} \: }}$

TO FIND :–

• Value of 'a' and 'b' = ?

SOLUTION :–

$\\ \: \implies \: { \bold{ \dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \: \: = a + b \sqrt{5} \: }}$

• Rationalization of denominator –

$\\ \: \implies \: { \bold{ \dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \times \dfrac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} } \: \: = a + b \sqrt{5} \: }}$

$\\ \: \implies \: { \bold{ \dfrac{(4 + 3 \sqrt{5}) {}^{2} }{(4 - 3 \sqrt{5} )( 4 + 3 \sqrt{5}) } = a + b \sqrt{5} \: }}$

• Using property –

$\\ \: \: \longrightarrow \: \: \large { \bold{ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab }}$

$\\ \: \: \longrightarrow \: \: \large { \bold{ {(a + b)}(a - b) = {a}^{2} - {b}^{2} }}$

• So that –

$\\ \: \implies \: { \bold{ \dfrac{16 + 45 + 24 \sqrt{5} }{(4 ) {}^{2} - (3 \sqrt{5} ) {}^{2} } = a + b \sqrt{5} \: }}$

$\\ \: \implies \: { \bold{ \dfrac{16 + 45 + 24 \sqrt{5} }{16 - 45 } = a + b \sqrt{5} \: }}$

$\\ \: \implies \: { \bold{ \dfrac{61+ 24 \sqrt{5} }{ - 29 } = a + b \sqrt{5} \: }}$

$\\ \: \implies \: { \bold{ \dfrac{61}{ - 29} + \dfrac{24 \sqrt{5} }{ - 29 } = a + b \sqrt{5} \: }}$

• Now compare –

$\\ \: \implies \: { \bold{a = - \dfrac{61}{ 29} \: \: and \: \: b = - \dfrac{24 }{ 29 } }}$

• Hence , $\: \: { \bold{a = - \dfrac{61}{ 29} \: \: and \: \: b = - \dfrac{24 }{ 29 } }}$

Answer 2

Given :-

$\rightarrowtail \tt \dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } = a + b \sqrt{5}$

Required to find :-

Values of " a " and " b " ?

Solution :-

Given :-

$\rightarrowtail \tt \dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } = a + b \sqrt{5}$

we need to find the values of ' a ' and ' b '

So,

Consider the LHS part ;

$\mapsto \sf \dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} }$

Here we need to factorise the denominator .

So,

Rationalising factor of 4 - 3√5 = 4 + 3√5

So,

$\mapsto \rm \dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \times \dfrac{4 + 3 \sqrt{5} }{ 4+ 3 \sqrt{5} }$

Here we need to use some identities ;

The identities are ;

• 1. ( x + y )² = x² + 2xy + y²

• 2. ( x + y ) ( x - y ) = x² - y²

• 3. ( x + y ) ( x + y ) = ( x + y )²

Using 2nd , 3rd identities solve it ;

So,

$\mapsto \rm \dfrac{(4 + 3 \sqrt{5} {)}^{2} }{(4 {)}^{2} - ( 3\sqrt{5} {)}^{2} }$

Now, using 1st identity expand the numerator

$\mapsto \rm \dfrac{(4 {)}^{2} + (3 \sqrt{5} {)}^{2} + 2(4)(3 \sqrt{5}) }{16 - 45}$

$\mapsto \rm \dfrac{16 + 45 + 24 \sqrt{5} }{ - 29}$

$\mapsto \rm \: \dfrac{61 + 24 \sqrt{5} }{ - 29}$

This implies ;

$\mapsto \rm \dfrac{ \: \: \: \: \: 61}{ -29} + \bigg( - \dfrac{ 24 \sqrt{5} }{ 29} \bigg)$

From this we can conclude that ;

The LHS is in the form of RHS

So,

Equal the values on both sides ,

$\mapsto \rm \dfrac{ \: \: \: \: \: 61}{ -29} + \bigg( - \dfrac{ 24 \sqrt{5} }{ 29} \bigg) = a + b \sqrt{5}$

Hence,

$\huge \leadsto \tt \: a \: = \: \frac{ \: \: \: \: 61}{ - 29}$

$\huge \leadsto \tt \: b \: = \: \frac{ \: \: \: \: \ 24}{ - 29}$