Question

HELP!!


`\sf If \: {x}^(2) + \frac{1}{ {x}^(2) } = 47, \: and \: x > 0, \: then \: what \: is \: value \: of \: x + (1)/(x)`

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Answer 1

`\qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star`

• x + 1/x = 7

`\textsf{ \underline{\underline{Steps to solve the problem} }:}`

`\qquad❖ \: \sf \: {x}^(2) + \cfrac{1}{ {x}^(2) } = 47`

`\qquad❖ \: \sf \: {x}^(2) + \cfrac{1}{ {x}^(2) } + 2 - 2 = 47`

( adding and subtracting 2, doesn't change the value )

`\qquad❖ \: \sf \: {x}^(2) + \cfrac{1}{ {x}^(2) } + 2 = 47 + 2`

`\qquad❖ \: \sf \: {x}^(2) + \cfrac{1}{ {x}^(2) } + \bigg(2 \sdot x \sdot \cfrac{1}{x} \bigg) = 49`

( x cancel out, so no change in value )

`\qquad❖ \: \sf \: \bigg(x + \cfrac{1}{x} \bigg) {}^(2) = 49`

( use identity a² + 2ab + b² = (a + b)² )

`\qquad❖ \: \sf \: \bigg(x + \cfrac{1}{x} \bigg) {}^{} = √(49)`

`\qquad❖ \: \sf \: x + \cfrac{1}{x} {}^{} = \pm7`

since we have to take positive value, i.e greater than 0

`\qquad❖ \: \sf \: x + \cfrac{1}{x} {}^{} = 7`

`\qquad \large \sf {Conclusion} :`

Therefore, the required value is 7

Answer 2

• 
  `x+\cfrac{1}{x} =7`

================

Given:

• 
  `x^2+\cfrac{1}{x^2} =47`

Add 2 to both sides of equation:

• 
  `x^2+\cfrac{1}{x^2}+2 =47+2`

Then follow the steps:

• 
  `x^2+\cfrac{1}{x^2}+2*x*\cfrac{1}{x} =49`

• 
  `(x+\cfrac{1}{x})^2=7^2`

Take square root of both sides to get:

• 
  `x+\cfrac{1}{x} =\pm\ 7`

We are taking the positive value as we are told x > 0, hence the answer is 7.