Question

if xsquare + 1 upon xsquare =79 find the value of x +1upon x

Answer 1

${x}^{2} + \frac{1}{ {x}^{2} } = 79 \\ \\ on \: adding \: 2 \: in \: both \: sides \\ \\ {(x)}^{2} + { (\frac{1}{x} )}^{2} + 2 = 79 + 2 \\ \\ {(x)}^{2} + {( \frac{1}{x} )}^{2} + 2(x)( \frac{1}{x} ) = 81 \\ \\ {(x + \frac{1}{x} )}^{2} = 81 \\ \\ x + \frac{1}{x} = 9$