Question

math
$given \: {x}^{2} + \frac{1}{ {x}^{2} } = 10 \\ x = what$
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Answer 1

Answer:

${x}^{2} + \frac{1}{ {x}^{2} } = 10 \: \: ......(i) \\ add \: 2 \: (or \: 2 \times x \times \frac{1}{x} ) \: to \: both \: side \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2x \times \frac{1}{x} = 10 + 2 \\ {(x + \frac{1}{x}) }^{2} = 12 \\ x + \frac{1}{x} = \frac{ + }{} \sqrt{12} = \frac{ + }{}2 \sqrt{3} \: \: ......(ii) \\ subract \: 2 \: (or \: 2x \times \frac{1}{x} ) \: from \: both \: side \: of \: eq(i) \\ {x}^{2} + \frac{1}{ {x}^{2} } - 2x \times \frac{1}{x} = 10 - 2 \\ (x - \frac{1}{x} ) {}^{2} = 8 \\ x - \frac{1}{x} = \frac{ + }{} \sqrt{8} = \frac{ + }{} 2 \sqrt{2} \: \: .......(iii) \\ add \: eq(ii) \: and \: (iii) \: we \: get \\ (x + \frac{1}{x} ) + (x - \frac{1}{x} ) =( \frac{ + }{} 2 \sqrt{3}) + \: ( \frac{ + }{} 2 \sqrt{2} ) \\ 2x = \frac{ + }{} 2 \sqrt{3} \frac{ + }{ } 2 \sqrt{2} \\ x = \frac{ \frac{ + }{} 2 \sqrt{3} \frac{ + }{ } 2 \sqrt{2}}{2} \\ there \: are \: four \: value \: of \: : \\ x _{1} = \frac{ + 2 \sqrt{3} + 2 \sqrt{2} }{2} = \sqrt{3} + \sqrt{2} \\ x _{2} = \frac{ - 2 \sqrt{3} - 2 \sqrt{2} }{?} = - ( \sqrt{3} + \sqrt{2} ) \\ x _{3} = \frac{2 \sqrt{3} - 2 \sqrt{2} }{2} = \sqrt{3} - \sqrt{2} \\ x _{4} = \frac{ - 2 \sqrt{3} + 2 \sqrt{2} }{2} = - \sqrt{3} + \sqrt{2}$

Answer 2

Answer:

${x}^{2} + \frac{1}{ {x}^{2} } = 10 \\ {x}^{2} + 1 = 10 \times {x}^{2} \\ {x}^{2} + 1 = 10 {x}^{2} \\ {x}^{2} - 10 {x}^{2} = - 1 \\ ( - 9 {x}^{2} ) = ( - 1) \\ ( - 3 {x})^{2} = ( { - 1})^{2} \\ - 3x = - 1 \\ 3x = 1$