Question

if
${x}^{2} + \frac{1}{ {x}^{2} } = 79find \: the \: value \: of \: x + \frac{1}{x}$
​

Answer 1

Step-by-step explanation:

${x}^{2} + \frac{1}{ {x}^{2} } =7 9 \\ \frac{ {x}^{4} + 1 }{ {x}^{2} } = 79 \\ {x}^{4} + 1 - 79 {x}^{2} = 0 \\ {x}^{4} - 79 {x}^{2} + 1 = 0 \\ {x}^{2} ( {x}^{2} - 79) + 1 = 0 \\ {x}^{2} = 0 .....x = \sqrt{0} = 0 \\ {x}^{2} - 79 + 1 = 0 \\ {x}^{2} - 78 = 0 \\ x = \sqrt{78} = 8.8$