Question

plzz solve the given question

Answer 1

${x}^{2} + \frac{1}{ {x}^{2} } = {(x - \frac{1}{x}) }^{2} + 2$
Substituting this value in the above question,

$12 {(x - \frac{1}{x}) }^{2} (2) - 28(x - \frac{1}{x} ) - 9 = 0 \\ \\ 12 {(x - \frac{1}{x}) }^{2} - 28(x - \frac{1}{x} ) - 9 + 24 = 0 \\ \\ 12 {(x - \frac{1}{x}) }^{2} - 28(x - \frac{1}{x} ) + 15 = 0$
$[/tex] Now for the sake of simplicity, lets take, [tex]x - \frac{1}{x} = y$


$12 {y}^{2} - 28y + 15 = 0$
$12 {y}^{2} - 18y - 10y + 15 = 0 \\ \\ 6y(2y - 3) - 5(2y + 3) = 0 \\ (6y - 5)(2y - 3) = 0$


$6y - 5 = 0 \\ y = \frac{ 5}{6} \\ \\ \\ or \\ \\ 2y - 3 = 0 \\ y = \frac{3}{2}$


So ,

$x - \frac{1}{x} = \frac{5}{6} \: \: or \: \: \frac{3}{2}$


Taking 5/6 as the value,

$x - \frac{1}{x} = \frac{5}{6} \\ 6 {x}^{2} - 6 = 5x \\ 6 {x}^{2} - 5x - 6 = 0\\ 6 {x}^{2} - 9x + 4x - 6 = . \\ 3x(2x - 3) + 2(2x - 3) = 0 \\ (3x + 2)(2x - 3 )= 0 \\ \\ so \\ \\ x = \frac{ - 2}{3} \: or \frac{3}{2}$




And if 3/2 is taken as the value then,

$x - \frac{1}{x} = \frac{3}{2} \\ 2 {x}^{2} - 3x - 2 = 0 \\ 2 {x}^{2} - 4x - x - 2 = 0 \\ 2x(x - 2) - (x + 2) = 0 \\ (2x - 1)(x + 2) = 0 \\ \\ \\ therefore \\ \\ x = \frac{1}{2 } \: or \: 2$