Question

please answer this question fast​

Answer 1

Answer is=

a=0

b=1

hope its help you

Answer 2

$\large\underline{\sf{Given \:Question - }}$

$\sf \: If \: \dfrac{7 + 3 \sqrt{5} }{7 - 3 \sqrt{5} } = a + b \sqrt{5}, \: find \: the \: value \: of \: a \: and \: b.$

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

Given that,

$\rm :\longmapsto\: \dfrac{7 + 3 \sqrt{5} }{7 - 3 \sqrt{5} } = a + b \sqrt{5}$

On rationalizing the denominator, we get

$\rm :\longmapsto\: \dfrac{7 + 3 \sqrt{5} }{7 - 3 \sqrt{5} } \times \dfrac{7 + 3 \sqrt{5} }{7 + 3 \sqrt{5} } = a + b \sqrt{5}$

We know,

$\red{\boxed{ \rm \:(x + y)(x - y) = {x}^{2} - {y}^{2}}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{ {(7 + 3 \sqrt{5} )}^{2} }{ {7}^{2} - {(3 \sqrt{5} )}^{2} } = a + b \sqrt{5}$

We know,

$\red{\boxed{ \rm \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{ {7}^{2} + {(3 \sqrt{5} )}^{2} + 2 \times 7 \times 3 \sqrt{5} }{49 - 45} = a + b \sqrt{5}$

$\rm :\longmapsto\:\dfrac{49 + 45 + 42 \sqrt{5} }{4} = a + b \sqrt{5}$

$\rm :\longmapsto\:\dfrac{94 + 42 \sqrt{5} }{4} = a + b \sqrt{5}$

$\rm :\longmapsto\:\dfrac{2(47 + 21 \sqrt{5}) }{4} = a + b \sqrt{5}$

$\rm :\longmapsto\:\dfrac{47 + 21 \sqrt{5}}{2} = a + b \sqrt{5}$

$\rm :\longmapsto\:\dfrac{47}{2} + \dfrac{21}{2} \sqrt{5} = a + b \sqrt{5}$

So, on comparing, we get

$\rm :\longmapsto\:\boxed{ \bf \:a = \dfrac{47}{2}} \: \: and \: \: \boxed{ \bf \:b = \dfrac{21}{2} }$

Additional Information :-

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)