Question

If $\displaystyle{\:x\:-\:\dfrac{1}{x}\:=\:5}$, find the value of $\displaystyle{\:1\:.\:x^2\:+\:\dfrac{1}{x^2}}$$\displaystyle{\:2\:.\:x^4\:+\:\dfrac{1}{x^4}}$​

Answer 1

Given

• $\sf{x - \dfrac{1}{x} = 5}$

To find

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

• $\sf{x^4 + \dfrac{1}{x^4}}$

Solution

• $\sf{x - \dfrac{1}{x} = 5}$

$\underline{ \bf \red{Identity~to~ be~ used~ :-}}$

$\qquad\leadsto \sf \purple{(a - b)^2 = a^2 + b^2 - 2ab}$

Squaring both sides,

$: \implies \sf \gray{ \bigg(x - \dfrac{1}{x} \bigg)^{2} = (5)^{2} }$

$: \implies \sf \gray{( {x})^{2} + \bigg( \dfrac{1}{x} \bigg)^{2} - 2 \big(x \big) \bigg( \frac{1}{x} \bigg) = 25}$

$: \implies \sf \gray{( {x})^{2} + \bigg( \dfrac{1}{x} \bigg)^{2} - 2 \big( \not x \big) \bigg( \frac{1}{ \not x} \bigg) = 25}$

$: \implies \sf \gray{{x}^{2} + \dfrac{1}{x^{2}} - 2 = 25}$

$: \implies \sf \gray{{x}^{2} + \dfrac{1}{x^{2}} = 25 + 2}$

$: \implies \sf \gray{{x}^{2} + \dfrac{1}{x^{2}} = 27}$

$\longmapsto \underline {\boxed{{ \bf \red{{x}^{2} + \dfrac{1}{x^{2}} = 27}}}} \pink \bigstar$

Now, finding the value of $\sf{x^4 + \dfrac{1}{x^4}}$

$\sf{x^4 + \dfrac{1}{x^4}}$

Squaring both the sides,

$: \implies \sf \gray{ \bigg({x}^{2} + \dfrac{1}{x^{2}} \bigg)^{2} = \big(27 \big)^{2} }$

$\underline{ \bf \red{Identity~ to~ be~ used~ :-}}$

$\qquad\leadsto \sf \purple{(a + b)^2 = a^2 + b^2 + 2ab}$

$: \implies \sf \gray{ \big({x}^{2} \big)^{2} + \bigg(\dfrac{1}{x^{2}} \bigg)^{2} + 2 \big(x^{2} \big) \bigg( \dfrac{1}{x^{2}} \bigg) = \big(27 \big)^{2}}$

$: \implies \sf \gray{ \big({x}^{2} \big)^{2} + \bigg(\dfrac{1}{x^{2}} \bigg)^{2} + 2 \big( \not x^{2} \big) \bigg( \dfrac{1}{ \not x^{2}} \bigg) = \big(27 \big)^{2}}$

$: \implies \sf \gray{ {x}^{4} + \dfrac{1}{x^{4}} + 2 = 729}$

$: \implies \sf \gray{ {x}^{4} + \dfrac{1}{x^{4}} = 729 - 2}$

$: \implies \sf \gray{ {x}^{4} + \dfrac{1}{x^{4}} = 727}$

$\longmapsto \underline {\boxed{{ \bf \red{{x}^{4} + \dfrac{1}{x^{4}} = 727}}}} \pink \bigstar$

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: Answer:-}}}}}$

◉ GIVEN:-

$\displaystyle{\:x\:-\:\dfrac{1}{x}\:=\:5}$

◉ TO FIND:-

$\displaystyle{\:1.)\:x^2\:+\:\dfrac{1}{x^2}}$

$\displaystyle{\:2.)\:x^4\:+\:\dfrac{1}{x^4}}$

◉FORMULA USED:-

${\boxed{\underline{\tt{\purple{(a-b)^2 = a^2 + b^2 -2ab }}}}}$

${\boxed{\underline{\tt{\purple{(a + b)^2 = a^2 + b^2 + 2ab }}}}}$

◉SOLUTION:-

$\leadsto{ x - \: \frac{1}{x} = 5}$

Squaring both sides:-

$\leadsto \: {(x - \frac{1}{x}) }^{2} = 25$

$\leadsto \: {x}^{2} + \frac{1}{ {x}^{2} } - 2 \times \cancel{ x} \times \frac{1}{ \cancel{x}} = 25$

$\leadsto \: {x}^{2} + \frac{1}{ {x}^{2} } = 25 + 2$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: { \boxed{ \underline{ \sf{ \blue{ {x}^{2} + \frac{1}{ {x}^{2} } = 27}}}}}$

NOW, TO FIND:-

$\displaystyle{\:x^4\:+\:\dfrac{1}{x^4}}$

We will again do squaring of

$\bold{ {x}^{2} + \frac{1}{ {x}^{2} } = 27}$

$\leadsto \: {( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = {27}^{2}$

$\leadsto \: {x}^{4} + \frac{1}{ {x}^{4} } + 2 \: \cancel{ {x}^{2}} . \frac{1}{ \cancel{ {x}^{2}} } = 729$

$\leadsto \: {x}^{4} + \frac{1}{ {x}^{4} } = 729 - 2$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: { \boxed{ \underline{ \sf{ \blue{ {x}^{4} + \frac{1}{ {x}^{4} } = 727}}}}}$

HOPE IT HELPS