Math, asked by manvisingh2007213, 3 months ago


If x -1/x =√6 , then find the value of x2 +1/x2​

Answers

Answered by asdeepsingh777
0

Answer:

8

Step-by-step explanation:

given:

x -  \frac{1}{x \:}  =  \: \sqrt{6}

squaring both sides:

(x -  \frac{1}{x} ) ^{2}  = 6

 {x}^{2}  +  (\frac{1}{x} )^{2}  - 2(x)( \frac{1}{x} ) = 6

 {x}^{2}  + ( \frac{1}{x} )^{2}  = 6 + 2

= 8

Answered by snehitha2
3

Answer :

The required value of x² + 1/x² = 8

Step-by-step explanation :

Given :

\sf x-\dfrac{1}{x}=\sqrt{6}

To find :

\sf the \ value \ of \ x^2+\dfrac{1}{x^2}

Identity :

(a - b)² = a² + b² - 2ab

Solution :

 x - 1/x = √6

Squaring on both sides,

   \sf \longrightarrow \bigg (x-\dfrac{1}{x}\bigg)^2=(\sqrt{6})^2 \\\\ \sf \longrightarrow  (x)^2+\bigg(\dfrac{1}{x}\bigg)^2-2(x)\bigg(\dfrac{1}{x}\bigg)=6 \\\\ \sf \longrightarrow x^2+\dfrac{1}{x^2}-2=6 \\\\ \sf \longrightarrow  x^2+\dfrac{1}{x^2}=6+2 \\\\ \sf \longrightarrow x^2+\dfrac{1}{x^2}=8

The required value of x² + 1/x² = 8

_______________________

 Some algebraic identities :

  (a + b)² = a² + 2ab + b²

  a² – b² = (a + b)(a – b)

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

  (a + b)³ = a³ + b³ + 3ab (a + b)

  (a – b)³ = a³ - b³ – 3ab (a – b)

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

 

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