![\huge\color{pink}\boxed{\colorbox{Black}{♔︎Answer♔︎}}](https://img.qammunity.org/2023/formulas/mathematics/college/7s3hm54uju4c6d9i828gf9rgsc0ukhwkbt.png)
To rationalise:-
![\frac{5}{ \sqrt{3 - √(5) } }](https://img.qammunity.org/2023/formulas/mathematics/high-school/11vd2va5umyrgbog58ao9tefslz2o50lwd.png)
There is a formula in math if there is root in denominator
for example
![(1)/( √(a - b) )](https://img.qammunity.org/2023/formulas/mathematics/high-school/vml6mawgi54eeh8tl1i1uj31foq43xua7k.png)
we can say rationalize by multiplying √(a-b) in numerator and denominator both
![(1)/( √(a - b) ) * ( √(a - b) )/( √(a - b) ) \\ ( √(a - b) )/(a - b)](https://img.qammunity.org/2023/formulas/mathematics/high-school/tyfbf8v0ymsd7obul7fy1tn74dvzw7v8m3.png)
In here
![\frac{5}{ \sqrt{3 - √(5) } } * \frac{ \sqrt{3 - √(5) } }{ \sqrt{3 - { √(5) } } } \\ \frac{5( \sqrt{3 - √(5) }) }{3 - √(5) }](https://img.qammunity.org/2023/formulas/mathematics/high-school/lvjo8h74ghly2manulzb1wldu5meks7x7l.png)
but still here is root to remove this we have to multiply
3 + √5 in numerator and denominator.
![\frac{5( \sqrt{3 - √(5) }) }{3 - √(5) } * (3 + √(5) )/(3 + √(5) ) \\ \frac{5( \sqrt{3 - √(5) })(3 + √(5) ) }{ {3}^(2) - { (√(5) )}^(2) } \\ \frac{5( \sqrt{3 - √(5) })(3 + √(5)) }{9 - 5} \\ \frac{5( \sqrt{3 - √(5) } )(3 + √(5)) }{4}](https://img.qammunity.org/2023/formulas/mathematics/high-school/e7mrr4omlovyqg1dg8wzvry686av1ne80a.png)