Question

HELLO FREIND

VIPUN HERE

⤵️HERE IS YOUR QUESTION ⤵️

$if \: {x}^{2} + \frac{1}{ {x}^{2} } = 27. \: find \: the \: value \: of \: each \: of \: the \: following \: \\ x + \frac{1}{x} \\ x - \frac{1}{x}$
⬆️12 points ⬆️

GIVE CORRECT ANSWER WITH CORRECT EXPLAINATION ..​

Answer 1

$\huge\rm\color{indigo}{Solution}$

$\sf{\underline{Given} \:: \: {x}^{2} + \frac{1}{ {x}^{2} } = 27 }$

$\sf{\underline{To \: find }\: : \: (i)x + \frac{1}{x} \: \: (ii)x - \frac{1}{x} }$

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

$\sf {(x + \frac{1}{x} )}^{2} = 27 + 2$

$\sf {(x + \frac{1}{x} )}^{2} = 29$

$\fbox{\sf x + \frac{1}{x} = \pm\sqrt{29} }$

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

$\sf {(x - \frac{1}{x}) }^{2} = 27 - 2$

$\sf {(x - \frac{1}{x}) }^{2} = 25$

$\sf x - \frac{1}{x} = \pm\sqrt{25}$

$\fbox{\sf x - \frac{1}{x} = \pm5 }$

Answer 2

\huge\rm\color{indigo}{Solution}Solution

\sf{\underline{Given} \:: \: {x}^{2} + \frac{1}{ {x}^{2} } = 27 }

Given

:x

2

+

x

2

1

=27

\sf{\underline{To \: find }\: : \: (i)x + \frac{1}{x} \: \: (ii)x - \frac{1}{x} }

Tofind

:(i)x+

x

1

(ii)x−

x

1

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

x

1

)

2

=x

2

+

x

2

1

+2

\sf {(x + \frac{1}{x} )}^{2} = 27 + 2(x+

x

1

)

2

=27+2

\sf {(x + \frac{1}{x} )}^{2} = 29(x+

x

1

)

2

=29

\fbox{\sf x + \frac{1}{x} = \pm\sqrt{29} }

x+

x

1

=±

29

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

x

1

)

2

=x

2

+

x

2

1

−2

\sf {(x - \frac{1}{x}) }^{2} = 27 - 2(x−

x

1

)

2

=27−2

\sf {(x - \frac{1}{x}) }^{2} = 25(x−

x

1

)

2

=25

\sf x - \frac{1}{x} = \pm\sqrt{25}x−

x

1

=±

25

\fbox{\sf x - \frac{1}{x} = \pm5 }

x−

x

1

=±5