Question

If a = (√3 + √ 2)/(√3 - √2) and b = (√3 - √2)/(√3 + √2) , find the value of a^2 − b^2 + ab .
PLS ANS ASAP!! TIA!!!

Answer 1

$\begin{gathered}a = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ on \: rationalizing \: we \: get \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} })^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ a = 5 + 2 \sqrt{6} \\ \\ b = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ using \: the \: identities \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} - 2 \times \sqrt{3} \times \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ = 5 - 2 \sqrt{6} \\ \\ {a}^{2} + {b}^{2} \\ \\ = {(5 + 2 \sqrt{6} )}^{2} + (5 - 2 \sqrt{6} )^{2} \\ \\ = ( {(5)}^{2} + {(2 \sqrt{6}) }^{2} + 2 \times 5 \times 2 \sqrt{6} ) + ( {(5)}^{2} + {(2 \sqrt{6} }^{2} ) - 2 \times 5 \times 2 \sqrt{6} ) \\ \\ = (25 + 24 + 20 \sqrt{6} ) + (25 + 24 - 20 \sqrt{6} ) \\ \\ = (49 + 20 \sqrt{6} ) + (49 - 20 \sqrt{6} ) \\ \\ = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} \\ \\ \end{gathered}$

= 49 + 49

= 98