Question

Find a and b if 3-√5/ 3+2 √5= a√5 -b​

Answer 1

Answer :

$\bull \: \sf a = \dfrac{9}{11}$

$\bull \sf \: b = \dfrac{19}{11}$

Explanation :

Given,

$\sf \to \: \dfrac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } = a \sqrt{5 } - b$

Taking L. H. S.

$\to \sf \dfrac{ 3 - \sqrt{5}}{3 + 2 \sqrt{5} }$

Rationalising the denominator,

$\sf \to \: \dfrac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } \times \dfrac{3 - 2 \sqrt{5} }{3 - 2 \sqrt{5} } \\ \sf \to \: \dfrac{(3 - \sqrt{5})(3 - 2\sqrt{5} )}{ {(3)}^{2} - {(2 \sqrt{5}) }^{2} } \\ \sf \to \: \dfrac{9 - 3 \sqrt{5} - 6 \sqrt{5} + 10 }{9 - 20} \\ \sf \to \: \dfrac{1 9- 9 \sqrt{5} }{ - 11} \\ \sf \to \: \dfrac{-19 + 9\sqrt{5}}{11}$

On comparing with R. H. S.

$\sf \implies \: \dfrac{-19}{ 11} + \dfrac{9}{11} \sqrt{5} = a \sqrt{5} - b \\ \therefore\sf a = \dfrac{9}{11} \: and \: b = \dfrac{19}{11}$