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

Step-by-step explanation:

Please only one brainlist answer

hope it helps you

Answer 2

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

$find \: {x}^{2} + \dfrac{1}{ {x}^{2} }$

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

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Identity to use : (a-b)² = a² - 2ab + b²

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

${5}^{2} = {x}^{2} - 2 + \dfrac{1}{ {x}^{2} }$

(x and 1/x gets cancelled)

$25 = {x}^{2} + \dfrac{1}{ {x}^{2} } - 2$

$25 + 2 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

$27 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

again applying the (a+b)² = a² + 2ab + b² identity, we get :

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

x² and 1/x² gets cancelled

${27}^{2} = {x}^{4} + 2 + \dfrac{1}{ {x}^{4} }$

$729 = {x}^{4} + \dfrac{1}{ {x}^{4} } + 2$

$729 - 2 = {x}^{4} + \dfrac{1}{ {x}^{4} }$

$727 = {x}^{4} + \dfrac{1}{ {x}^{4} }$

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Some more identities : Extras

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

(a+b)³ = a³ + 3a²b + 3ab² + b³

(a-b)³ = a³ - 3a²b + 3ab² - b³

a³ - b³ = (a-b)(a²+ab+b²)

a³ + b³ = (a+b)(a²-ab+b²)

a³ + b³ + c³ = (a+b+c)(a²+b²+c²-ab-bc-ca)

Conditional identity:

if a+b+c = 0,

a³ + b³ + c³ = 3abc

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