Question

$\frac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7}$
a = ? , b = ?​

Answer 1

Given :

$\sf \: \dfrac{3 + \sqrt{7} }{3 - \sqrt{7} } = a \: + \: b \sqrt{7}$

To find :

• Value of a & b

Method used :

• Firstly rationalising denominator on LHS, and then comparing it with RHS

Identity used :

• (a+b)² = a² + b² + 2ab
• (a+b)(a-b) = a² - b²

Solution :

LHS =>

Multiply and divide by (3+√7)

we get,

$\dfrac{(3 + \sqrt{7 } ) \times (3 + \sqrt{7})} {(3 - \sqrt{7} ) \times (3 + \sqrt{7} )} \\ \\ = \dfrac{ {(3 + \sqrt{7}) }^{2} }{( {3}^{2}) - ( { \sqrt{7} )}^{2} } \\ \\ = \dfrac{ ({3})^{2} + ( { \sqrt{7} )}^{2} + (2 \times 3 \times \sqrt{7} ) }{9 - 7} \\ \\ = \dfrac{9 + 7 + \: 6 \sqrt{7} }{2} \\ \\ = \dfrac{16 + 6 \sqrt{7} }{2} \\ \\ = \dfrac{16}{2} + \dfrac{6 \sqrt{7} }{2} \\ \\ = 8 + 3 \sqrt{7}$

Also , RHS = a + b√7

we know that LHS = RHS

=> 8 + 3√7 = a + b√7

on comparing

we get,

a = 8

b = 3

ANSWER :

• a = 8
• b = 3