Question

simplfy (4+√5/4-√5. )+ (4-√5/4+√45) rationalise method​

Answer 1

Appropriate Question -

simplfy $\sf \: \frac{4 + \sqrt{5} }{4 - \sqrt{{5}}}$ + $\sf \: \frac{4 - \sqrt{5}}{4 + \sqrt{\bf{5}}}$ by rationalise method.

Given -

• $\sf \: \dfrac{4 + \sqrt{5} }{4 - \sqrt{{5}}}$

• $\sf \: \dfrac{4 - \sqrt{5} }{4 + \sqrt{5} }$

To -

• Simplify

Solution -

In the question, we are provided with the 2 fractions and we need to simplify it. For that, First we will rationalise both the denominators and then we will add both the obtained values, that will give us the answer.

For Rationalising -

$\sf \:\dfrac{numerator}{denominator}\: \times \dfrac{denominator}{denominator}$

Let -

$\sf \: \dfrac{4 + \sqrt{5} }{4 - \sqrt{{5}}}$ = a

$\sf \: \dfrac{4 - \sqrt{5} }{4 + \sqrt{5} }$ = b

On substituting the values -

$\sf \: \longrightarrow \dfrac{4 + \sqrt{5} }{4 - \sqrt{5} } \: \times \dfrac{4 + \sqrt{5} }{4 + \sqrt{5} } \\ \\ \sf \longrightarrow \: \dfrac{(4 + \sqrt{5})^{2} }{ {(4)}^{2} - \sqrt{(5})^{2}} \\ \\ \sf \longrightarrow \: \dfrac{ {(4)}^{2} + { \sqrt{(5})^{2} + 2(4)( \sqrt{5)}}}{16 - 5} \\ \\ \sf \longrightarrow \: \dfrac{16 + 5 + 8 \sqrt{5} }{11} \\ \\ \sf \longrightarrow \: \dfrac{21 + 8 \sqrt{5} }{11}$

Similarly -

$\sf \longrightarrow \: \dfrac{4 - \sqrt{5} }{4 + \sqrt{5}} \\ \\ \sf \longrightarrow \: \dfrac{4 - \sqrt{5} }{4 + \sqrt{5}} \: \times \frac{4 - \sqrt{5} }{4 - \sqrt{5}} \\ \\ \sf \longrightarrow \: \dfrac{(4 - \sqrt{5})^{2} }{ {(4)}^{2} - {( \sqrt{5})}^{2} } \\ \\ \sf \longrightarrow \: \dfrac{ {(4)}^{2} + {( \sqrt{5})^{2} - 2(4)( \sqrt{5})}}{16 - 5} \\ \\ \sf \longrightarrow \: \dfrac{16 + 5 - 8 \sqrt{5} }{11} \\ \\ \sf \longrightarrow \: \dfrac{21 - 8 \sqrt{5} }{11}$

At the end -

$\sf \dfrac{21 + 8 \sqrt{5}}{11} \: + \dfrac{21 - 8 \sqrt{5} }{11} \\ \\ \sf \: (taking \: 11 \: as \: common) \\ \\ \sf \: \dfrac{(21 + \cancel{8 \sqrt{5}}) \: + (21 - \cancel{8 \sqrt{5})}}{11} \\ \\ \sf \: \frac{21 + 21}{11} \\ \\ \sf \: \frac{42}{11} \: \underline{\: ans}$

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Answer 2

Given :

• $\sf \dfrac{4 + \sqrt{5} }{4 - \sqrt{5} } \: , \: \dfrac{4 - \sqrt{5} }{4 + \sqrt{5} }$

$\: \: \:$

$\: \: \:$

To Find :

Simplfy

• $\sf \dfrac{4 + \sqrt{5} }{4 - \sqrt{5} } \: + \: \dfrac{4 - \sqrt{5} }{4 + \sqrt{5} }$

$\: \: \:$

$\: \: \:$

Solution :

Here in the question We are given to simplify the Two fractions. So, Firstly we will rationalise the denominator of both the fractions. As in the Question it is to simplify using rationalise Method.After rationalising the denominators we can easily add the both Fractions.

$\: \: \:$

Rationalising The denominator :

$\: \: \:$

$\bf : \implies \dfrac{4 + \sqrt{5} }{4 - \sqrt{5} }$

$\: \: \:$

$\sf : \implies \dfrac{4 + \sqrt{5} }{4 - \sqrt{5} } \times \dfrac{4 + \sqrt{5} }{4 + \sqrt{5} }$

$\: \: \:$

$\sf : \implies \dfrac{(4 + \sqrt{5})^{2} }{( {4})^{2} - \sqrt{(5)^{2} } }$

$\: \: \:$

$\sf: \implies \dfrac{( {4})^{2} + \sqrt{( {5})^{2} } + 8( \sqrt{5} )}{16 - 5}$

$\: \: \:$

$\sf : \implies \dfrac{16 + 5 + 8 \sqrt{5} }{11}$

$\: \: \:$

$\bf : \implies {\red {\dfrac{21 + 8 \sqrt{5} }{11}}} \: \: taking \: as \: (i)$

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$\: \: \:$

$\bf: \implies\dfrac{4 - \sqrt{5} }{4 + \sqrt{5} }$

$\: \: \:$

$\sf: \implies\dfrac{4 - \sqrt{5} }{4 + \sqrt{5} } \times \dfrac{4 - \sqrt{5} }{4 - \sqrt{5} }$

$\: \: \:$

$\sf : \implies \dfrac{(4 - \sqrt{5})^{2} }{( {4})^{2} - \sqrt{(5)^{2} } }$

$\: \: \:$

$\sf: \implies \dfrac{( {4})^{2} + \sqrt{( {5})^{2} } - 8( \sqrt{5} )}{16 - 5}$

$\: \: \:$

$\sf : \implies \dfrac{16 + 5 - 8 \sqrt{5} }{11}$

$\: \: \:$

$\bf : \implies {\red{\dfrac{21 + 8 \sqrt{5} }{11}}} \: \: taking \: as \: (ii)$

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$\: \: \:$

Now, Adding (i) and (ii)

$\: \: \:$

$\sf : \implies \dfrac{21 + 8 \sqrt{5} }{11} + \dfrac{21 - 8 \sqrt{5}}{11}$

$\: \: \:$

$\sf \implies \dfrac{(21 + \cancel{8 \sqrt{5}}) \: + (21 - \cancel{8 \sqrt{5})}}{11}$

$\: \: \:$

$\sf: \implies \dfrac{21 + 21}{11}$

$\: \: \:$

$: \implies{ \boxed{\red{ \bf {\dfrac{42}{11} }}}}\: \pink\star$

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