Question

factorise the following

Answer 1

$hi...$

${x}^{2} + \frac{1}{ {x}^{2} } - 11 = 0$

$(multiply \: by \: {x}^{2} )$

${x}^{4} + 1 - 11 {x}^{2} = 0$

${x}^{4} - 11 {x}^{2} + 1 = 0$

$let \: {x}^{2} \: be \: y$

${y}^{2} - 11y + 1 = 0$

$comparing \: with \\ \: a {y}^{2} + by + c = 0$

a = 1 , b = -11 , c = 1
${b}^{2} - 4ac = 117$

By the formula method...

$y = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} or \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

$y = \frac{ - ( - 11) + \sqrt{117} }{2(1)} or \frac{ - ( - 11) - \sqrt{117} }{2(1)}$

$y = \frac{11 + \sqrt{117} }{2} or \: \frac{11 - \sqrt{117} }{2}$

Re substituting
$y = {x}^{2}$

${x}^{2} = \frac{11 + \sqrt{117} }{2 } or \: \frac{11 - \sqrt{117} }{2}$

By taking square root to both sides

$x = \sqrt{ \frac{11 + \sqrt{117} }{2} } \: or \sqrt{ \frac{11 - \sqrt{117} }{2} }$

$hope \: this \: helps$