Question

if a =√3-√2by√3+√2and b=√3+√2by√3-√2find the value of a^2+b^2-5ab

Answer 1

Hey friend,
Here is the answer you were looking for:
$a = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

On rationalizing the denominator we get,

$a = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

Using the identities:

${(a - b)}^{2} = {a}^{2} + { b }^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$

$a = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} - 2( \sqrt{3})( \sqrt{2} ) }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ a = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ a = 5 - 2 \sqrt{6}$

$b = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

On rationalizing the denominator we get,

$b = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

Using the identities:

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$

$b = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} +2 ( \sqrt{3} )( \sqrt{2}) }{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } \\ \\ b = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ b = 5 + 2 \sqrt{6}$

Now,

${a}^{2} + {b}^{2} - 5ab$
Putting the values we get,

${(5 - 2 \sqrt{6}) }^{2} + {(5 + 2 \sqrt{6} )}^{2} - 5(5 - 2 \sqrt{6} )(5 + 2 \sqrt{6} )$

Using same identities

$( {(5)}^{2} + {(2 \sqrt{6} )}^{2} - 2(5)(2 \sqrt{6} )) + ( {(5)}^{2} + {(2 \sqrt{6} )}^{2} + 2(5)(2 \sqrt{6} )) - 5( {(5)}^{2} - {(2 \sqrt{6} )}^{2} ) \\ \\ = (25 + 24 - 20 \sqrt{6} ) + (25 + 24 + 20 \sqrt{6} ) + 5(25 - 24) \\ \\ = 49 - 20 \sqrt{6} + 49 + 20 \sqrt{6} - 5(1) \\ \\ = 98 - 5 \\ \\ = 93$
Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
☺☺