Question

HELP!!
$\sf If \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 47, \: and \: x > 0, \: then \: what \: is \: value \: of \: x + \dfrac{1}{x}$

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Answer 1

Answer:

• $x+\cfrac{1}{x} =7$

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

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.

Answer 2

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

• 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 )

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

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

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