Question

plz solve it and tell me step by step​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{ \sqrt{3} + 1}{ \sqrt{3} - 1} = a + b \sqrt{3}$

On rationalizing the denominator, we get

$\rm :\longmapsto\:\dfrac{ \sqrt{3} + 1}{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } = a + b \sqrt{3}$

$\rm :\longmapsto\:\dfrac{( \sqrt{3} + 1) ^{2} }{ (\sqrt{3} - 1)( \sqrt{3} + 1) } = a + b \sqrt{3}$

$\rm :\longmapsto\: \dfrac{3 + 1 + 2 \sqrt{3} }{ {( \sqrt{3} )}^{2} - {(1)}^{2} } = a + b \sqrt{3}$

$\: \: \: \: \: \: \: \red{\bigg \{ \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy\bigg \}}$

$\: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm :\longmapsto\: \dfrac{4+ 2 \sqrt{3} }{3 - 1} = a + b \sqrt{3}$

$\rm :\longmapsto\: \dfrac{2(2+ \sqrt{3})}{2} = a + b \sqrt{3}$

$\rm :\longmapsto\: \blue{2} + \red{\sqrt{3}} = \blue{a} + \red{ b \sqrt{3} }$

So, On comparing we get,

$\bf\implies \:a = 2 \: \: \: and \: \: \: b = 1$

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

Answer:

${ \large{ \pmb{ \sf{★ Given... }}}}$

${ \sf{ \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1} = a + b \sqrt{3} }}$

${ \large{ \pmb { \sf{★To \: Find... </p><p>}}}}$

Value of a and b..?

${ \large{ \pmb{ \sf{★Solution... }}}}$

$\implies{ \sf{ \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} }}$

Now Rationalizing the Denominator,

${ \implies{ \sf{ \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1 }{ \sqrt{3} + 1 } }}}$

$\: { \implies{ \sf{ \frac{ {( \sqrt{3} + 1 )}^{2} }{ {( \sqrt{3} )}^{2} - {1}^{2} } }}}$

$\:{ \implies{ \sf{ { \frac{({ \sqrt{3} )}^{2} + {1}^{2} + 2( \sqrt{3} )(1)}{3 - 1} }}}}$

${ \implies{ \sf{ \frac{3 + 1 + 2 \sqrt{3} }{2} }}}$

$\: { \implies{ \sf{ \frac{4 + 2 \sqrt{3} }{2} }}}$

$\: { \implies{ \sf{ \frac{2(2 + \sqrt{ 3} )}{2} }}}$

$\: { \implies{ \sf{ \frac{ \cancel{2}(2 + \sqrt{ 3} )}{ \cancel{2}} }}}$

${ \implies{ \sf{2 + \sqrt{3} }}}$

Now Comparing 2 + √3 with a + b√3

We get,

• a = 2

• b = 1

Therefore,

• Value of a = 2
• Value of b = 1

Used Formula:

• (a + b) (a - b) = a² - b²

• (a + b)² = a² + b² + 2ab