Question

Plz solve the question​

Answer 1

Given :-

$x - \dfrac{1}{x} = 3$

To find :-

${x}^{2} + \dfrac{1}{ {x}^{2}} \: and \: {x}^{4} + \dfrac{1}{ {x}^{4} }$

Identity to use :-

(a-b)² = a² - 2ab + b²

$(x - \dfrac{1}{x})^{2} = (x)^{2} - 2(x)( \dfrac{1}{x}) + (\dfrac{1}{ {x}})^{2}$

Substituting the values,

$({3})^{2} = {x}^{2} - 2( \not x)( \dfrac{1}{ \not x}) + \dfrac{1}{ {x}^{2} }$

$9 = {x}^{2} - 2 + \dfrac{1}{ {x}^{2} }$

$9 + 2 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

$11 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

Identity to use :-

(a+b)² = a² + 2ab + b²

$( {x}^{2} + \dfrac{1}{ {x}^{2} })^{2} = ( {x}^{2})^{2} + 2(x)( \dfrac{1}{x}) + (\dfrac{1}{ {x}^{2} })^{2}$

Substituting the values,

${11}^{2} = {x}^{4} + 2( \not {x}^{2} )( \dfrac{1}{ \not {x}^{2} }) + \dfrac{1}{ {x}^{4} }$

$121= {x}^{4} + 2 + \dfrac{1}{ {x}^{4} }$

$121 - 2= {x}^{4} + \dfrac{1}{ {x}^{4} }$

$119 = {x}^{4} + \dfrac{1}{ {x}^{4} }$

Some more identities :-

a²-b² = (a+b)(a-b)

(x+a)(x+b) = x² + x(a+b) + ab

(a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

Answer 2

GIVEN :-

$\\ \sf \: x - \dfrac{1}{x} = 3$

TO FIND :-

$\\ \sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: and \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{4} + \dfrac{1}{ {x}^{4} }$

IDENTITY USED :-

• (a-b)² = a² + b² - 2ab

• (a+b)² = a² + b² + 2ab

SOLUTION :-

We have ,

$\\ \underline{ \underline{ \sf \: x - \dfrac{1}{x} = 3}}$

Squaring both sides ,

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

We know , (a-b)² = a² + b² - 2ab

Here ,

• a = x
• b = 1/x

Substituting values we get ,

$\\ \implies \sf \: {x}^{2} + ( { \dfrac{1}{x}) }^{2} - 2( \cancel{x})( \dfrac{1}{ \cancel {x}} ) = 9\\ \\ \implies \sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = 9 \\ \\ \implies \sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 9 + 2 \\ \\ \implies \underline{ \boxed{ \bf \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 11 } }$

Now we will square both sides in the above equation.

$\\ \\ \implies \sf \: ( {x}^{2} + { \dfrac{1}{ {x}^{2} } )}^{2} = {11}^{2}$

We know , (a+b)² = a² + b² + 2ab

Here ,

• a = x²
• b = 1/x²

Substituting values we get...

$\\ \implies \sf \: {( {x}^{2} )}^{2} + {( \dfrac{1}{ {x}^{2} } )}^{2} + 2( \cancel{ {x}^{2} })( \dfrac{1}{ \cancel{{x}^{2}} } ) = 121 \\ \\ \implies \sf \: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 = 121 \\ \\ \implies \sf \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 121 - 2 \\ \\ \implies \underline{ \boxed{\bf \: {x}^{4} + \frac{1}{ {x}^{4} } = 119 }}$

Hence ,

$\\ \boxed{ \sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 11 } \: \: \: \: \: \: \sf \: and \: \: \: \: \: \: \boxed { \sf \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 119 }$

MORE IDENTITIES :-

• (a+b)(a-b) = a² - b²
• (a+x)(a+y) = a² + (x+y)a + xy
• (a+b)³ = a³ + 3a²b + 3ab² + b³
• (a-b)³ = a³ - 3a²b + 3ab² - b³