Question

please solve question 5 and 6​

Answer 1

hope it helps you mate....

Answer 2

$\star\:\:\:\sf\large\underline\blue{Question\:5:-}$

Given,

$\sf x + \frac{1}{x} = 12$

Squaring both sides, we get,

$\sf{(x + \frac{1}{x} )}^{2} = {12}^{2}$

$\sf \implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times \cancel{x }\times \frac{1}{ \cancel{x}} = 144$

$\sf \implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 144$

Subtracting 2 from both sides, we get,

$\sf \implies {x}^{2} + \frac{1}{ {x}^{2} } = 142$

Therefore,

$\boxed{ \sf {x}^{2} + \frac{1}{ {x}^{2} } = 142}$

$\star\:\:\:\sf\large\underline\blue{Question\:6:-}$

Given,

$\sf x + \frac{1}{x} = 5$

Squaring both sides, we get,

$\sf{(x + \frac{1}{x} )}^{2} = {5}^{2}$

$\sf \implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times \cancel{x }\times \frac{1}{ \cancel{x}} = 25$

$\sf \implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 25$

Subtracting 2 from both sides, we get,

$\sf \implies {x}^{2} + \frac{1}{ {x}^{2} } = 23$

Therefore,

$\boxed{ \sf {x}^{2} + \frac{1}{ {x}^{2} } = 23}$