Question

simplify 3+ root 7/3-root7= a+b root7 find the value of a and b

Answer 1

HELLO.....FRIEND!!

THE ANSWER IS HERE,
$= > \: \frac{3 + \sqrt{7} }{3 - \sqrt{7} }$

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

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

$= > \: \frac{9 + 7 + 6 \sqrt{7} }{2}$

$= > \: \frac{16 + 6 \sqrt{7} }{2}$

$= > \: 8 + 3 \sqrt{7} .$

From the question,

$= > \: 8 + 3 \sqrt{7} = a + b \sqrt{7}$

$= > \: a = 8 \:$
$= > \: b = 3.$

:-)Hope it helps u.

Answer 2

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

By Rationalization,

$\frac{3 + \sqrt{7} }{3 - \sqrt{7} } \times \frac{3 + \sqrt{7} }{3 + \sqrt{7} } = a + b \sqrt{7}$

$\frac{(3 + \sqrt{7} )^{2} }{(3)^{2} - ( \sqrt{7} ) ^{2} } = a + b \sqrt{7}$

$\frac{9 + 7 + 6 \sqrt{7} }{9 - 7} = a + \sqrt{7}$
$\frac{16 + 6 \sqrt{7} }{2} = a + b \sqrt{7}$


$\frac{8 + 3 \sqrt{7} }{1} = a + b \sqrt{7}$


$8 + 3 \sqrt{7} = a + b \sqrt{7}$


a = 8

b = 3



I hope this will help you


-by ABHAY