Question

please answer this as early as possible ​

Answer 1

Question :–

• Determine the rational numbers 'a' and 'b' if :

$\\ { \bold{ \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } + \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } = a + b \sqrt{3} }}$

ANSWER :–

GIVEN :–

$\\ \: \: { \huge{.}} \: \: { \bold{ \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } + \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } = a + b \sqrt{3} }}$

TO FIND :–

• Value of 'a' and 'b' = ?

SOLUTION :–

$\\ \implies { \bold{ \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } + \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } = a + b \sqrt{3} }}$

• We should write this as –

$\\ \implies { \bold{ \dfrac{( \sqrt{3} - 1)^{2} + {( \sqrt{3} + 1) }^{2} }{( \sqrt{3} + 1 )( \sqrt{3} - 1)}= a + b \sqrt{3} }}$

▪︎ Using identities –

$\\ \dashrightarrow { \bold{ {(a \: \pm \: b)}^{2} = {a}^{2} + {b}^{2} \: \pm \: 2ab }}$

$\\ \dashrightarrow { \bold{ {(a + b)}(a - b) = {a}^{2} - {b}^{2} }}$

▪︎ So that –

$\\ \implies { \bold{ \dfrac{( \sqrt{3} )^{2} + 1 - 2 \sqrt{3} + {( \sqrt{3} ) }^{2} + 1 + 2 \sqrt{3} }{( \sqrt{3})^{2} - (1)^{2} }= a + b \sqrt{3} }}$

$\\ \implies { \bold{ \dfrac{3 + 1 + 3 + 1}{3 - 1}= a + b \sqrt{3} }}$

$\\ \implies { \bold{ \dfrac{8}{2}= a + b \sqrt{3} }}$

$\\ \implies { \bold{4= a + b \sqrt{3} }}$

▪︎ Now Let's compare –

$\\ \dashrightarrow \large \:\: { \boxed{ \bold{a = 4}}}$

• And –

$\\ \dashrightarrow \large \:\: { \boxed{ \bold{b = 0}}}$