Question

If 2/root 3 +root 5 add 5/root 3 - root 5 = a root3 + b root5, find a and b.​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\large{\boxed{ \tt a = - \dfrac{7}{2} , \: b = - \dfrac{3}{2} }}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :-

$\tt \dfrac{2}{ \sqrt{3} + \sqrt{5}} + \dfrac{5}{ \sqrt{3} - \sqrt{5}} = a \sqrt{3} + b \sqrt{5}$

To find :- Values of a and b

Solution :-

$\tt \dfrac{2}{ \sqrt{3} + \sqrt{5}} + \dfrac{5}{ \sqrt{3} - \sqrt{5}} = a \sqrt{3} + b \sqrt{5}$

Consider Left Hand Side of the equation

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

Now rationalise the denominators of each term in the above expression

$\tt = (\dfrac{2}{ \sqrt{3} + \sqrt{5}} \times \dfrac{ \sqrt{3} - \sqrt{5} }{ \sqrt{3} - \sqrt{5} })+ (\dfrac{5}{ \sqrt{3} - \sqrt{5}} \times \dfrac{ \sqrt{3} + \sqrt{5} }{ \sqrt{5} + \sqrt{5} })$

$\tt = \dfrac{2( \sqrt{3} - \sqrt{5})}{ {( \sqrt{3})}^{2} - {( \sqrt{5})}^{2} } + \dfrac{5( \sqrt{3} + \sqrt{5})}{ {( \sqrt{3})}^{2} - {( \sqrt{5})}^{2} }$

[Since (x + y)(x - y) = x² - y² ]

$\tt = \dfrac{2( \sqrt{3} - \sqrt{5})}{3 - 5} + \dfrac{5 \sqrt{3} + 5\sqrt{5} }{3 - 5}$

$\tt = \dfrac{2( \sqrt{3} - \sqrt{5})}{ - 2} + \dfrac{5 \sqrt{3} + 5 \sqrt{5} }{ - 2}$

$\tt = \dfrac{2\sqrt{3} - 2\sqrt{5}}{ - 2} + \dfrac{5 \sqrt{3} + 5 \sqrt{5} }{ - 2}$

$\tt = \dfrac{2 \sqrt{3} - 2 \sqrt{5} + (5 \sqrt{3} + 5 \sqrt{5})}{ - 2}$

$\tt = \dfrac{2 \sqrt{3} - 2 \sqrt{5} + 5 \sqrt{3} + 5 \sqrt{5}}{ - 2}$

$\tt = \dfrac{7 \sqrt{3} + 3 \sqrt{5} }{ - 2}$

$\tt = \dfrac{7 \sqrt{3} }{ - 2} + \dfrac{3 \sqrt{5} }{ - 2}$

$\tt = - \dfrac{7 \sqrt{3} }{2} + ( - \dfrac{3 \sqrt{5} }{2})$

$\tt = - \dfrac{7 \sqrt{3} }{2}- \dfrac{3 \sqrt{5} }{2}$

Now consider

$\bf \dfrac{2}{ \sqrt{3} + \sqrt{5}} + \dfrac{5}{ \sqrt{3} - \sqrt{5}} = a \sqrt{3} + b \sqrt{5}$

$\tt - \dfrac{7 \sqrt{3} }{2}- \dfrac{3 \sqrt{5} }{2} = a \sqrt{3} + b \sqrt{5}$

Comparing on both sides

$\tt a \sqrt{3} = - \dfrac{7 \sqrt{3} }{2}$

$\tt a\cancel{\sqrt{3}} = - \dfrac{7\cancel{ \sqrt{3}} }{2}$

$\tt a = - \dfrac{7}{2}$

$\tt b \sqrt{5} = - \dfrac{3 \sqrt{5} }{2}$

$\tt b\cancel{ \sqrt{5}}= - \dfrac{3\cancel{\sqrt{5}} }{2}$

$\tt b = - \dfrac{3}{2}$

$\Huge{\boxed{ \tt a = - \dfrac{7}{2} , \: b = - \dfrac{3}{2} }}$