Question

root5 +root3 by root5 - root 3 = а- b root15​

Answer 1

Question :

$\sf \small \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a - b \sqrt{15}$

Process :

• First we have to rationalize the denominator on the left hand side

• Then we need to simplify them

• And at last we need to compare both the sides to get the value of a and b

Solution :

$\sf \small \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a - b \sqrt{15}$

Using formula (a + b)(a – b) = a² – b² to get the denominator

Here,

• a² → (√5)²

• b² → (√3)²

$\sf \small \rightarrow \frac{ (\sqrt{5} + \sqrt{3})( \sqrt{5} + \sqrt{3} ) }{( \sqrt{5} - \sqrt{3} )( \sqrt{5} + \sqrt{3} )} = a - b \sqrt{15}$

$\sf \small \rightarrow \frac{( \sqrt{5} + \sqrt{3})^{2} }{ {( \sqrt{5} )}^{2} - {( \sqrt{3}) }^{2} } = a - b \sqrt{15}$

Using formula (a+b)² = a² + b² + 2ab to get the numerator

Here,

• a² → (√5)²

• b² → (√3)²

• 2ab → 2•√5•√3

$\sf \small \rightarrow \frac{( \sqrt{5} )^{2} + 2 \times \sqrt{5} \times \sqrt{3} + {( \sqrt{3}) }^{2} }{5 - 3} = a - b \sqrt{15}$

$\sf \small \rightarrow \frac{5 + 2 \sqrt{15} + 3}{2} = a - b \sqrt{15}$

$\sf \small \rightarrow \frac{5 + 3 + 2 \sqrt{15} }{2} = a - b \sqrt{15}$

$\sf \small \rightarrow \frac{8 + 2 \sqrt{15} }{2} = a - b \sqrt{15}$

$\sf \small \rightarrow \frac{2(4 + \sqrt{15}) }{2} = a - b \sqrt{15}$

$\sf \small \rightarrow 4 + \sqrt{15} = a - b \sqrt{15}$

Comparing both sides we get

• a → 4

• b → -1