Question

Rationalise the denominator of 5/√(3-√5)
Pls send the answer by today​

Answer 1

Answer:

$\dfrac{5(3+\sqrt{5})\sqrt{3-\sqrt{5}}}{4}$

$\textsf{or}\quad \dfrac{5\sqrt{3+\sqrt{5}}}{2}$

Step-by-step explanation:

$\textsf{Given expression}:\dfrac{5}{\sqrt{3-\sqrt{5}}}$

Method 1

$\textsf{Multiply by the conjugate}\quad \dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{3-\sqrt{5}}}:$

$\implies \dfrac{5}{\sqrt{3-\sqrt{5}}} \times \dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{3-\sqrt{5}}}=\dfrac{5\sqrt{3-\sqrt{5}}}{(\sqrt{3-\sqrt{5}})(\sqrt{3-\sqrt{5}})}$

Simplify the denominator using the radical rule $\sqrt{a} \sqrt{a} =a$:

$\implies (\sqrt{3-\sqrt{5}})(\sqrt{3-\sqrt{5}})=3-\sqrt{5}$

Therefore:

$\implies \dfrac{5\sqrt{3-\sqrt{5}}}{(\sqrt{3-\sqrt{5}})(\sqrt{3-\sqrt{5}})}= \dfrac{5\sqrt{3-\sqrt{5}}}{3-\sqrt{5}}$

$\textsf{Multiply by the conjugate}\quad \dfrac{3+\sqrt{5}}{3+\sqrt{5}}:$

$\implies \dfrac{5\sqrt{3-\sqrt{5}}}{3-\sqrt{5}} \times \dfrac{3+\sqrt{5}}{3+\sqrt{5}}=\dfrac{5\sqrt{3-\sqrt{5}}(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}$

Simplify the denominator:

$\implies (3-\sqrt{5})(3+\sqrt{5})=9+3\sqrt{5}-3\sqrt{5}-5=4$

Therefore:

$\implies \dfrac{5(3+\sqrt{5})\sqrt{3-\sqrt{5}}}{4}$

Method 2

$\textsf{Multiply by the conjugate}\quad \dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{3+\sqrt{5}}}:$

$\implies \dfrac{5}{\sqrt{3-\sqrt{5}}} \times \dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{3+\sqrt{5}}}=\dfrac{5\sqrt{3+\sqrt{5}}}{(\sqrt{3-\sqrt{5}})(\sqrt{3+\sqrt{5}})}$

Simplify the denominator using the radical rule $\sqrt{a} \sqrt{b} =\sqrt{ab}$:

$\implies (\sqrt{3-\sqrt{5}})(\sqrt{3+\sqrt{5}})=\sqrt{(3-\sqrt{5})(3+\sqrt{5})$

$\implies\sqrt{(3-\sqrt{5})(3+\sqrt{5})}=\sqrt{9-5}=\sqrt{4}=2$

Therefore:

$\implies \dfrac{5\sqrt{3+\sqrt{5}}}{2}$

Answer 2

$\huge\color{pink}\boxed{\colorbox{Black}{♔︎Answer♔︎}}$

To rationalise:-

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

There is a formula in math if there is root in denominator

for example

$\frac{1}{ \sqrt{a - b} }$

we can say rationalize by multiplying √(a-b) in numerator and denominator both

$\frac{1}{ \sqrt{a - b} } \times \frac{ \sqrt{a - b} }{ \sqrt{a - b} } \\ \frac{ \sqrt{a - b} }{a - b}$

In here

$\frac{5}{ \sqrt{3 - \sqrt{5} } } \times \frac{ \sqrt{3 - \sqrt{5} } }{ \sqrt{3 - { \sqrt{5} } } } \\ \frac{5( \sqrt{3 - \sqrt{5} }) }{3 - \sqrt{5} }$

but still here is root to remove this we have to multiply

3 + √5 in numerator and denominator.

$\frac{5( \sqrt{3 - \sqrt{5} }) }{3 - \sqrt{5} } \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} } \\ \frac{5( \sqrt{3 - \sqrt{5} })(3 + \sqrt{5} ) }{ {3}^{2} - { (\sqrt{5} )}^{2} } \\ \frac{5( \sqrt{3 - \sqrt{5} })(3 + \sqrt{5}) }{9 - 5} \\ \frac{5( \sqrt{3 - \sqrt{5} } )(3 + \sqrt{5}) }{4}$