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

User Squidbe
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2 Answers

8 votes
8 votes


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

To rationalise:-


\frac{5}{ \sqrt{3 - √(5) } }

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

for example


(1)/( √(a - b) )

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


(1)/( √(a - b) ) * ( √(a - b) )/( √(a - b) ) \\ ( √(a - b) )/(a - b)

In here


\frac{5}{ \sqrt{3 - √(5) } } * \frac{ \sqrt{3 - √(5) } }{ \sqrt{3 - { √(5) } } } \\ \frac{5( \sqrt{3 - √(5) }) }{3 - √(5) }

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}

User BigLex
by
2.9k points
24 votes
24 votes

Answer:


\frac{5(3+√(5))\sqrt{3-√(5)}}{4}


\textsf{or}\quad \frac{5\sqrt{3+√(5)}}{2}

Explanation:


\textsf{Given expression}:\frac{5}{\sqrt{3-√(5)}}

Method 1


\textsf{Multiply by the conjugate}\quad \frac{\sqrt{3-√(5)}}{\sqrt{3-√(5)}}:


\implies \frac{5}{\sqrt{3-√(5)}} * \frac{\sqrt{3-√(5)}}{\sqrt{3-√(5)}}=\frac{5\sqrt{3-√(5)}}{(\sqrt{3-√(5)})(\sqrt{3-√(5)})}

Simplify the denominator using the radical rule
√(a) √(a) =a:


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

Therefore:


\implies \frac{5\sqrt{3-√(5)}}{(\sqrt{3-√(5)})(\sqrt{3-√(5)})}= \frac{5\sqrt{3-√(5)}}{3-√(5)}


\textsf{Multiply by the conjugate}\quad (3+√(5))/(3+√(5)):


\implies \frac{5\sqrt{3-√(5)}}{3-√(5)} * (3+√(5))/(3+√(5))=\frac{5\sqrt{3-√(5)}(3+√(5))}{(3-√(5))(3+√(5))}

Simplify the denominator:


\implies (3-√(5))(3+√(5))=9+3√(5)-3√(5)-5=4

Therefore:


\implies \frac{5(3+√(5))\sqrt{3-√(5)}}{4}

Method 2


\textsf{Multiply by the conjugate}\quad \frac{\sqrt{3+√(5)}}{\sqrt{3+√(5)}}:


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

Simplify the denominator using the radical rule
√(a) √(b) =√(ab):


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


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

Therefore:


\implies \frac{5\sqrt{3+√(5)}}{2}