Question

Kindly answer this question correctly

Answer 1

Correct question is

$\rm \red{ \huge{Question}}$

$\rm \implies \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a + b \sqrt{15} \: \: then \: \: show \: \: that \: \: a = 4 \: \: and \: \: b = 1$

$\rm \red{ \huge{Solution}}$

$\rm \implies \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a + b \sqrt{15}$

Take

$\rm \implies \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

Using rationalization methods

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

Using this identity

$\rm(a + b) = {a}^{2} + {b}^{2} + 2ab$

$\rm \: (a - b)(a + b) = {a}^{2} - {b}^{2}$

$\implies \rm \: \dfrac{ (\sqrt{5} + \sqrt{3} ) {}^{2} }{( \sqrt{5} ) {}^{2} - (\sqrt{ 3}) {}^{2} }$

$\rm \implies \dfrac{( \sqrt{5} ) {}^{2} + (\sqrt{ 3} ) {}^{2} + 2 \times \sqrt{5} + \sqrt{3} }{5 - 3}$

$\rm \implies \dfrac{5 + 3 + 2 \sqrt{15} }{2}$

$\rm \implies \dfrac{8 + 2 \sqrt{15} }{2}$

$\rm \: \implies4 + \sqrt{15}$

hence proved

a = 4 and b = 1

Answer 2

Step-by-step explanation:

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Answer♡

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