Question

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Answer 1

${\tt{\underline{Given :-}}}$

$\bullet\begin{cases} \bf \dfrac{3-2\sqrt{5}}{6-\sqrt{5}} = a+b\sqrt{5}\\ \bf \; a\;\;\&\;\;b \;are\;rational\end{cases}$

${\tt{\underline{To\;find :-}}}$

The respective values of a and b

${\tt{\underline{Answer :-}}}$

On rationalizing the denominator of the fraction

LHS

$\tt\dfrac{3-2\sqrt{5}}{6-\sqrt{5}}$

$\tt\dfrac{3-2\sqrt{5}}{6-\sqrt{5}}\times\dfrac{6-\sqrt{5}}{6-\sqrt{5}}$

$\tt\dfrac{3-2\sqrt{5}\times 6-\sqrt{5}}{6-\sqrt{5}\times6-\sqrt{5}}$

$\tt\dfrac{18+3\sqrt{5}-12\sqrt{5}-10}{36-\sqrt{5}\times\sqrt{5}}$

$\tt\dfrac{18+3\sqrt{5}-12\sqrt{5}-10}{36-5}$

$\tt\dfrac{8-9\sqrt{5}}{31}$

RHS

$\tt a+b\sqrt{5}$

On comparing both sides we have got

a = 8/31 & b = -9/31

Answer 2

Answer:

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