Question

If q=√2+√3, find the values of q-1/q and q2+1/q2​

Answer 1

Given -

$q = \sqrt{2} + \sqrt{3}$

To find -

$1)q - \frac{1}{q}$

$2) {q}^{2} + \frac{1}{ {q}^{2} }$

Solution -

$q = \sqrt{2} + \sqrt{3}$

$\frac{1}{q} = \frac{1}{ \sqrt{2} + \sqrt{3} }$

Rationalising the denominator we will get -

$\frac{1}{q} = \frac{ \sqrt{2} - \sqrt{3} }{( \sqrt{2} + \sqrt{3} )( \sqrt{2} - \sqrt{3} ) }$

$\frac{1}{q} = \frac{ \sqrt{2} - \sqrt{3} }{ {( \sqrt{2}) }^{2} - { (\sqrt{3}) }^{2} }$

$\frac{1}{q} = \frac{ \sqrt{2} - \sqrt{3} }{2 - 3}$

$\frac{1}{q} = \frac{ \sqrt{2} - \sqrt{3} }{ - 1}$

$\frac{1}{q} = - \sqrt{2 } + \sqrt{3}$

Now,

$= q - \frac{1}{q}$

Putting the values we will get -

$= \sqrt{2} + \sqrt{3} - ( - \sqrt{2} + \sqrt{3} )$

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

$= 2 \sqrt{2}$

$\boxed{\bf \: \therefore \: The \: value \: of \: q - \frac{1}{q} \: is \: 2 \sqrt{2} }$

Again,

$= {q}^{2} + \frac{1}{ {q}^{2} }$

Putting the values we will get -

$= {( \sqrt{2} + \sqrt{3} )}^{2} + {( \sqrt{3} - \sqrt{2}) }^{2}$

$= ( { \sqrt{2} )}^{2} +( { \sqrt{3}) }^{2} + 2 \times \sqrt{2} \times \sqrt{6} + {( \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} - 2 \times \sqrt{3 } \times \sqrt{2}$

$= 2 + 3 + 2 \sqrt{6} + 3 + 2 - 2 \sqrt{6}$

$= 2 + 3 + 3 + 2$

$= 5 + 5$

$= 10$

$\boxed{\bf \: \therefore \: The \: value \: of \: {q}^{2} + \frac{1}{ {q}^{2} } is \: 10}$