Question

If x^2+(1/x)^2=23,find the value of x-1/x​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

given :

${x}^{2} + \dfrac{1}{ {x}^{2} } = 23$

to find :

$x - \frac{1}{x}$

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

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

x and 1/x gets cancelled in -2(x)(1/x)

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

by rearranging,

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

substituting the value of x² + 1/x² as 23

$( {x - \frac{1}{x}) }^{2} = 23 - 2$

$( {x - \dfrac{1}{x} })^{2} = 21$

$( {x - \dfrac{1}{x} }) = {21}^{ \frac{1}{2} }$

hence the value of x + 1/x = √21

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Verification :

By substituting the values,

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

substituting x-1/x for √21

$(21^{ \frac{1}{2} }) ^{2} = {x}^{2} - 2 + \dfrac{1}{ {x}^{2} }$

$21 = {x}^{2} - 2 + \cfrac{1}{ {x}^{2} }$

$21 + 2 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

$23 = {x}^{2} + \dfrac{1}{ {x}^{2} }$

Therefore verified.

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Some more identities :

(a+b)² = a² + 2ab + b²

a²-b² = (a+b)(a-b)

(x+a)(x+b) = x² + x(a+b) + (ab)

(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)

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

a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)

Conditional identity:

a³ + b³ c³ = 3abc if a+b+c = 0

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