Question

please give explanation of this q with answer​

Answer 1

Question:-

$\\ \sf{If \: x = 3 + 2 \sqrt{2} , then \: the \: value of \: {x}^{2} - \dfrac{1}{ {x}^{2} } \: is \: equal \: to - }$

Required Answer:-

Given:-

$\sf{\rightarrow{x = 3 + 2 \sqrt{2}}}$

To Find:-

$\sf{\rightarrow{The \: value \: of \: {x}^{2} - \dfrac{1}{ {x}^{2} }}}$

Solution:-

We were given:-

$\sf{x = 3 + 2 \sqrt{2}}$

Therefore,

$\sf{\dfrac{1}{x} = \dfrac{1}{3 + 2 \sqrt{2}}}$

$\\ \sf{\implies{\dfrac{1}{x} = \frac{1 \times 3 - 2 \sqrt{2} }{(3 + 2 \sqrt{2})(3 - 2 \sqrt{2})}} }$

$\\ \sf{\implies{ \dfrac{1}{x} = \dfrac{3 + 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} }}}$

$\\ \sf{\implies{ \dfrac{1}{x} = \dfrac{3 + 2 \sqrt{2} }{9 - 8}}}$

$\\ \sf{\implies{ \dfrac{1}{x} = \bold{3 + 2 \sqrt{2}}}}$

Hence,

$\\ \sf{x + \dfrac{1}{x} = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2}}$

$\\ \sf{\implies{x + \dfrac{1}{x} =\bold{ 6}}}\blue{ - - - - - - - - - - - - - - (1)}$

Again,

$\sf{x - \dfrac{1}{x} =3 + 2 \sqrt{2} - (3 - 2 \sqrt{2})}$

$\\ \sf{\implies{x - \dfrac{1}{x} = 3 + 2 \sqrt{2} - 3 + 2 \sqrt{2}}}$

$\\ \sf{\implies{x - \dfrac{1}{x} =\bold{ 4 \sqrt{2}}}} \blue{ - - - - - - - - - - - - - - (2)}$

Now,

$\\ \sf{\bold{ {x}^{2} - \dfrac{1}{ {x}^{2}}}}$

$\\ \sf{\implies{ {(x)}^{2} - \dfrac{1}{ {(x)}^{2}}} }$

$\\ \sf{\implies{\big(x + \dfrac{1}{x}\big)\big (x - \dfrac{1}{x} \big)}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\rm{ {a}^{2} - {b}^{2} = (a + b)(a - b)}}$

Substituting the values from equation 1 and 2:-

$\\ \sf{\implies{6\times 4 \sqrt{2}}}$

$\\\sf{\implies{\bold{\red{24 \sqrt{2}}}}}$

$\\ \underline{\boxed{\rm{Hence, \: the \: value \: of \: {x}^{2} - \dfrac{1}{ {x}^{2} } \: is \: \bold{24 \sqrt{2}}}}}$