Question

√5+√3/√5-√3=a+b√15
$to a or b ka man bateye$
​

Answer 1

We have,

• To determine value of a & b, if

$\green{ \tt \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } =a + b \sqrt{15} }$

Here,

• On equating the rationalised LHS with given RHS, we can figure out a & b

Thus,

• Rationalising the LHS,

$\to \: \tt \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \\ \\ \to \: \tt \frac{(\sqrt{5} + \sqrt{3}) \times (\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3}) \times (\sqrt{5} + \sqrt{3})}$

◇ If, numbers are in form of $\bf(a + b)(a + b),$then, identity used is

$\boxed{ \bullet \: \bf \green{(a + b)(a + b) } = \pink{{(a + b)}^{2} }}$

◇ And, if numbers are in form of $\bf(a + b)(a - b)$, then, identity used is

$\boxed{ \bullet \: \bf \green{(a + b)(a - b) } = \pink{ {a}^{2} - {b}^{2} }}$

Therefore,

$\to \: \tt \frac{ {( \sqrt{5} + \sqrt{3} )}^{2} }{ {( \sqrt{5} )}^{2} - {( \sqrt{3} )}^{2} } \\ \\ \small \to \: \tt \frac{ {( \sqrt{5} )}^{2} + {( \sqrt{3} )}^{2} + 2( \sqrt{5} )( \sqrt{3} )}{5 - 3} \\ \\ \to \: \tt \frac{5 + 3 + 2 \sqrt{15} }{2} \\ \\ \to \: \tt \frac{8 + 2 \sqrt{15} }{2} \: \: = \frac{8}{2} + \frac{2 \sqrt{15} }{2} \\ \\ \to \: \sf4 + \sqrt{15}$

NOW,

• Equating obtained (simplified) value of LHS with RHS, we get,

$\tt{}a + b \sqrt{15} = 4 + \sqrt{15} \\ \\ \therefore \: \: \boxed{ \bf{}a = \red4 }\: \: \: \: \: \: \: \: \& \: \: \: \: \: \: \: \: \boxed{ \bf{}b = \red1}$

Thus,

• Value of a is 4.
• Value of b is 1.