Question

if √5+√3/√5-√3=a+b√15 then find the value of a and b

Answer 1

√5+√3/√5-√3×√5-√3/√5-√3
= √5(√5-√3)+√3(√5-√3)/(√5)^2-(√3)^2
= 5-√15+√15-3/2
= 2/2=1
it is the answer

Answer 2

Here's your answer!!

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$= > \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a + b \sqrt{15}$

We will first rationalise it ,

$= > \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} }$

$= > \frac{ ({ \sqrt{5} + \sqrt{3}) }^{2} }{ {( \sqrt{5}) - {( \sqrt{3}) }^{2} } } = a + b \sqrt{15}$

$= > \frac{5 + 3 + 2 \sqrt{15} }{2} = a + b \sqrt{15}$

$= > \frac{8 + 2 \sqrt{15} }{2} = a + b \sqrt{15}$

$= > \frac{8}{2} + \frac{2 \sqrt{15} }{2} = a + b \sqrt{15}$

Now,


$= > a = \frac{8}{2} = 4$

$= > b \sqrt{15} = \frac{2 \sqrt{15} }{2}$

$\sqrt{15 } \: got \: cancelled \: from \: both \: side$

$= > b = \frac{2}{2} = 1$

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Hope it helps you!! :)