Question

$if x=4 \sqrt{6} \div \sqrt{2} + \sqrt{3}, find the value of x + 2 \sqrt{2} \div x - 2 \sqrt{2} + x + 2 \sqrt{3} \div x - 2 \sqrt{3}$
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Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\bf \: if \: x = \dfrac{4 \sqrt{6} }{ \sqrt{3} + \sqrt{2} } \: then \: find \: the \: value \: of$

$\bf \:  \dfrac{x + 2\sqrt{2} }{x - 2 \sqrt{2} } + \dfrac{x + 2 \sqrt{3} }{x - 2 \sqrt{3} }$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

Process of componendo and Dividendo

If the ratio of a to b is equal to the ratio of c to d,

then the ratio of a + b to a − b is equal to the ratio of c + d to c − d.

If a:b = c:d then (a+b):(a-b) = (c+d):(c-d).

This property is called the componendo and dividendo rule.

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$\large\underline\purple{\bold{Solution :- }}$

$\large\underline{\bold{❥︎Step :- 1 }}$

$\sf \:  ⟼x = \dfrac{4 \sqrt{6} }{ \sqrt{3} + \sqrt{2} }$

$\sf \:  ⟼x = \dfrac{2 \times 2 \times \sqrt{2} \times \sqrt{3} }{ \sqrt{3} + \sqrt{2} }$

$\sf \:  ⟼x = \dfrac{(2 \sqrt{2} ) \times (2 \sqrt{3}) }{ \sqrt{3} + \sqrt{2} }$

$\sf \:  ⟼\dfrac{x}{2 \sqrt{2} } = \dfrac{2 \sqrt{3} }{ \sqrt{3} + \sqrt{2} }$

☆ Apply Componendo and Dividendo, we get

$\sf \:  ⟼ \: \dfrac{x + 2 \sqrt{2} }{x - 2 \sqrt{2} } = \dfrac{2 \sqrt{3} + \sqrt{2} + \sqrt{3} }{2 \sqrt{3} - \sqrt{3} - \sqrt{2} }$

$\sf \:  ⟼ \: \dfrac{x + 2 \sqrt{2} }{x - 2 \sqrt{2} } = \dfrac{3 \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } - - - (1)$

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$\large\underline{\bold{❥︎Step :- 2 }}$

$\bf \:  ⟼ x = \dfrac{4 \sqrt{6} }{ \sqrt{3} + \sqrt{2} }$

$\sf \:  ⟼x = \dfrac{2 \times 2 \times \sqrt{2} \times \sqrt{3} }{ \sqrt{3} + \sqrt{2} }$

$\sf \:  ⟼x = \dfrac{(2 \sqrt{2} ) \times (2 \sqrt{3}) }{ \sqrt{3} + \sqrt{2} }$

$\sf \:  ⟼\dfrac{x}{2 \sqrt{3} } = \dfrac{2 \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

☆ Apply Componendo and Dividendo, we get

$\sf \:  ⟼\dfrac{x + 2 \sqrt{3} }{x - 2 \sqrt{3} } = \dfrac{2 \sqrt{2} + \sqrt{3} + \sqrt{2} }{2 \sqrt{2} - \sqrt{3} - \sqrt{2} }$

$\sf \:  ⟼\dfrac{x + 2 \sqrt{3} }{x - 2 \sqrt{3} } = \dfrac{3 \sqrt{2} + \sqrt{3} }{ \sqrt{2} - \sqrt{3} }$

$\sf \:  ⟼\dfrac{x + 2 \sqrt{3} }{x - 2 \sqrt{3} } = - \dfrac{3 \sqrt{2} + \sqrt{3} }{ \sqrt{3} - \sqrt{2} } - - - (2)$

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$\large\underline{\bold{❥︎Step :- 3 }}$

☆ Adding equation (1) and equation (2), we get

$\sf \:  ⟼\dfrac{x + 2 \sqrt{2} }{x - 2 \sqrt{2} } + \dfrac{x + 2 \sqrt{3} }{x - 2 \sqrt{3} }$

$\sf \:  ⟼ = \dfrac{3 \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } - \dfrac{3 \sqrt{2} + \sqrt{3} }{ \sqrt{3} - \sqrt{2} }$

$\sf \:  ⟼ = \dfrac{3 \sqrt{3} + \sqrt{2} - 3 \sqrt{2} - \sqrt{3} }{ \sqrt{3} - \sqrt{2} }$

$\sf \:  ⟼ = \dfrac{2 \sqrt{3} -2 \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$\sf \:  ⟼ = \dfrac{2 \cancel{( \sqrt{3} - \sqrt{2} ) }}{\cancel{( \sqrt{3} - \sqrt{2} ) }} = 2$

$\bf\implies \:\dfrac{x + 2 \sqrt{2} }{x - 2 \sqrt{2} } + \dfrac{x + 2 \sqrt{3} }{x - 2 \sqrt{3} } = 2$