Question

solve the problems ​

Answer 1

Solution :-

Given that ,

$= > x = \frac{2}{3 + \sqrt{7} }$

We know that rationalisation factor for 3+ √7 is 3- √7. So , we multiply numerator numerator and denominator by 3- √7 , to get :-

$= > x = \frac{2}{3 + \sqrt{7} } \times \frac{3 - \sqrt{7} }{3 - \sqrt{7} }$

$= > \frac{2(3 - \sqrt{7}) }{(3^{2}) - ( \sqrt{7})^{2} }$

$= > \frac{2 - 3 \sqrt{7} }{9 - \sqrt{7} }$

$= > 3 - \sqrt{7}$

Hence ,

$= > x - 3 = - \sqrt{7}$

On squaring both sides , we get :-

$= > (x - 3)^{2} = 7$

Hence , the Value of the given expression is (d).