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Alex_Xolod [135]
2 years ago
5

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

Mathematics
2 answers:
Maksim231197 [3]2 years ago
8 0

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}}}

<u>Method 1</u>

\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}

<u>Method 2</u>

\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}

Hunter-Best [27]2 years ago
7 0

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

<u>To</u><u> </u><u>rationalise</u><u>:</u><u>-</u>

\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}

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