Question

if
${x}^{2} + \frac{1}{ {x}^{2} } = 34$
find the value of
$x + \frac{1}{x}$
​

Answer 1

$\bold{\huge{\underline{\underline{\rm{ GiVen :}}}}}$

${x}^{2} + \frac{1}{ {x}^{2} } = 34$

$\huge{\underline{\underline{\purple{♡Answer→}}}}$

${x}^{2} + \frac{1}{ {x}^{2} } = 34 \\ \\ ( {x + \frac{1}{x}) }^{2} - 2 \times x \times \frac{1}{x} = 34 \\ \\ ( {x + \frac{1}{x}) }^{2} - 2 = 34 \\ \\ ( {x + \frac{1}{x}) }^{2} = 34 + 2 \\ \\ ( {x + \frac{1}{x}) }^{2} = 36 \\ \\ x + \frac{1}{x} = \sqrt{36} \\ \\ x + \frac{1}{x} = 6$