Question

Rationalise it 1\3+√5 please

Answer 1

not sure but according to me it's correct

Answer 2

$\mathfrak{\underline{\underline{\green{Answer:-}}}}$

$\mathsf{\dfrac{1}{3+ \sqrt{5}}= \dfrac{3- \sqrt{5}}{4} }$

$\mathfrak{\underline{\underline{\green{Explanation:}}}}$

Given:

$\mathsf{\dfrac{1}{3+ \sqrt{5}} }$

Solution:

The rationalising factor of $\mathsf{\dfrac{1}{3+ \sqrt{5}}}$is

$\mathsf{{3- \sqrt{5}} }$

By rationalising the denominator

$\mathsf{= \dfrac{1}{3+ \sqrt{5}} \times \dfrac{3- \sqrt{5}}{3- \sqrt{5}}}$

$\mathsf{= \dfrac{1}{(3+ \sqrt{5})(3- \sqrt{5})} }$

$\mathsf{= \dfrac{1}{{3}^{2}- {(\sqrt{5})}^{2}} }$

$\mathsf{= \dfrac{3- \sqrt{5}}{9-5} }$

$\mathsf{= \dfrac{3- \sqrt{5}}{4} }$

Hence,

$\mathsf{= \dfrac{3- \sqrt{5}}{4} }$