Question

if (x+1/x)=4,then the value of (x²+1/x²) is ​

Answer 1

$\hookrightarrow \bf \: Here \: is \: solution$

$x + \frac{1}{x} = 4 \\ \sf taking \: squares \: on \: both \: sides \\ {(x + \frac{1}{x})}^{2} = {4}^{2} \\ {x}^{2} + 2 \times x \times \frac{1}{x} + {( \frac{1}{x} )}^{2} = 16 \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \boxed{ {x}^{2} + \frac{1}{ {x}^{2} } = 14}$

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Answer 2

$\large{\underline{\underline{\pmb{\sf{Given:}}}}}$

$\sf{\dashrightarrow x+ \dfrac{1}{x} = 4}$

$\large{\underline{\underline{\pmb{\sf{To\:Find:}}}}}$

$\sf{\dashrightarrow x^2+ \dfrac{1}{x^2}}$

$\large{\underline{\underline{\pmb{\sf{Solution:}}}}}$

In this problem, the main identity applied is:

$\underline{\boxed{\sf{(a+b)^2 = \bf a^2+b^2+2ab}}}$

Similarly, we know that,

$\sf{\dashrightarrow (x+ \dfrac{1}{x})^2 = x^2+1/x^2+2 \times x \times \dfrac{1}{x}}$

$\sf{\dashrightarrow (4)^2 = x^2+1/x^2+2 \times \cancel{x} \times \dfrac{1}{\cancel{x}}}$

$\sf{\dashrightarrow 16 = x^2+1/x^2+2}$

$\sf{\dashrightarrow 16-2 = x^2+1/x^2}$

$\dashrightarrow{\boxed{\sf{x^2+ \dfrac{1}{x^2} = 14}}}$

Hence, solved.