\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} =\)

(sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3))

multiple both numerator and denominator by :
(sqrt(5)+sqrt(3))

gives:

{[(sqrt(5)+sqrt(3))] ^ 2} / 2

or

4 + sqrt (15)

Take conjugate on both sides. which gives.

(sqrt(5) + sqrt(3)) (sqrt(5) + sqrt(3)) / (sqrt(5) - sqrt(3)) (sqrt(5) + sqrt(3))

This gives (sqrt(5) +sqrt(3))^2 / (Sqrt(5)^2) - (sqrt(3))^2

=> 5 + 3 + 2 sqrt(15) / 2

=> 2(4 + sqrt(15))/2

=> 4 + sqrt(15)

Thanks

This type of questions can be solved by using formula

(a+b)(a-b) = a^2 - b^2

Denominator is in the form of a-b , hence we will multiply numerator and denominator by the form a +b .i.e. sqrt(5) + sqrt(3)

sqrt(5) + sqrt(3)/sqrt(5) - sqrt(3)

[sqrt(5) + sqrt(3)]*[sqrt(5) + sqrt(3)]/[sqrt(5) + sqrt(3)]*[sqrt(5) - sqrt(3)]

5+3 + 2*sqrt(5)*sqrt(3) / 5-3

8+2*sqrt(5)*sqrt(3)/2

4+sqrt(15)

Chets asked if 4+sqrt(5)*sqrt(3) = 1/(4-sqrt(5)*sqrt(3)) can be extrapolated into a broad generalization. No, it can't be!

Here's why.

Consider an expression \(1/(a-\sqrt{bc})\) [modeled on 1/(4-sqrt(5)*sqrt(3)) ]
Multiplying Numerator and Denominator with \((a+ \sqrt{bc})\), we get
\((a+ \sqrt{bc})/(a^2-bc)\)

So, the reason why 1/(4-sqrt(5)*sqrt(3) got simplified into 4+sqrt(5)*sqrt(3) was because the Denominator of the above expression is equal to 1 in this particular case (a=4, b=5, c=3. So,\(a^2-bc = 4^2-5*3 = 1\))