Question

If root 5+root 3/2 root 5-3 root 3 =a-b root 15, find 'a' and 'b'

Answer 1

$\huge{\underline{\bold{Solution:}}}$

Given :

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

Let's simplify L.H.S by rationalising it's denominator.

$\frac{ \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - 3 \sqrt{3} } \\ \\ \frac{ \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - 3 \sqrt{3} } \times \frac{2 \sqrt{5} + 3 \sqrt{3} }{2 \sqrt{5} + 3 \sqrt{3} } \\ \\ \frac{ \sqrt{5}(2 \sqrt{5} + 3 \sqrt{3} ) + \sqrt{3} (2 \sqrt{5} + 3 \sqrt{3} )}{(2 \sqrt{5}) {}^{2} - (3 \sqrt{3}) {}^{2} } \\ \\ \frac{10 + 3 \sqrt{15} + 2 \sqrt{15} + 9 }{20 - 27} \\ \\ \frac{19 + 5 \sqrt{15} }{ - 7}$

Now,

$\frac{ - 19}{7} + \frac{5 }{7} \sqrt{15}$

On comparing the result of L.H.S from R.H.S , we get ;

$\bold{a = \frac{ - 19}{7} \: \: and \: \: b = \frac{5}{7} }$


Thanks for the question!

Answer 2

First rationalize: root 5 + root 3 divided by root 5 - root 3,

we get, (root 5 + root 3)2 / (root 5 - root 3)( root 5 + root 3)

= (8+2 root 15)/2

now compare LHS and RHS,

(given), (8+2root15) / 2 = a+root 15b

=> 2(4+ root 15) / 2 = a+root15 b [here we removed the common factor and cancelled]

=> 4+ root 15 = a+ root 15 b

therefore, the value of 'a' can be 4 and the value of 'b' can be 1 ( as any no. multiplied with one will be the no. itself)

hope this helps u (cuz i went in detail!!)