Question

if a=1-√7÷1+√7 and b =1+√7÷1-√7 then find the value of a^2+ab+b^2

Answer 1

${ \boxed{ \boxed{a = \frac{1 - \sqrt{7} }{1 + \sqrt{7} } }}} \\ { \boxed{ \boxed{b = \frac{1 + \sqrt{7} }{1 - \sqrt{7} } }}}$

Now, Rationalising a & b, we get,

$a = \frac{ {(1 - \sqrt{7} )}^{2} }{(1 - \sqrt{7} )(1 + \sqrt{7} )} = \frac{1 + 7 - 2 \sqrt{7} }{1 - 7} \\ \\ \: \: \: = \frac{8 - 2 \sqrt{7} }{ - 6} = { \red{ \frac{2 \sqrt{7} - 8}{6} }}$

$b = \frac{ {(1 + \sqrt{7} )}^{2} }{(1 + \sqrt{7} )(1 - \sqrt{7} )} = \frac{1 + 7 + 2 \sqrt{7} }{1 - 7} \\ \\ \: \: \: = \frac{8 + 2 \sqrt{7} }{ - 6} = { \red{\frac{ - 8 - 2 \sqrt{7} }{6} }}$

Now,

${a}^{2} + {b}^{2} + ab = {(a + b)}^{2} - ab \\ \\ = > ( \frac{ - 8 - 2 \sqrt{7} }{6} + \frac{ - 8 + 2\sqrt{7} }{6} )^{2} - (\frac{ - 8 - 2 \sqrt{7} }{6})(\frac{ - 8 + 2\sqrt{7} }{6}) \\ \\ = > ( \frac{ - 8 - 2 \sqrt{7} - 8 + 2 \sqrt{7} }{6} )^{2} - ( \frac{ {( - 8)}^{2} - {(2 \sqrt{7} )}^{2} }{36} ) \\ \\ = > {( \frac{ - 16}{6} )}^{2} - ( \frac{64 - 28}{36} ) \\ \\ = > \frac{256}{36} - \frac{36}{36} \\ \\ = > \frac{256 - 36}{36} = \frac{220}{36}$

Hence,

${ \boxed{ \huge { \red{{a}^{2} + {b}^{2} + ab = \frac{220}{36} }}}}$