Question

Please silve it fast ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\:9\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) - 9\bigg(x + \dfrac{1}{x} \bigg) - 52 = 0 - - (1)$

To solve this equation, Let assume that,

$\red{\rm :\longmapsto\:x + \dfrac{1}{x} = y - - - (2)}$

On squaring both sides, we get

$\red{\rm :\longmapsto\:{x}^{2} + \dfrac{1}{ {x}^{2}}+ 2 \times x \times \dfrac{1}{x} = {y}^{2} }$

$\red{\rm :\longmapsto\:{x}^{2} + \dfrac{1}{ {x}^{2}}+ 2 = {y}^{2} }$

$\red{\rm :\longmapsto\:{x}^{2} + \dfrac{1}{ {x}^{2}} = {y}^{2} - 2}$

On substituting all these values in equation (1), we get

$\rm :\longmapsto\:9( {y}^{2} - 2) - 9y - 52 = 0$

$\rm :\longmapsto\:9{y}^{2} - 18 - 9y - 52 = 0$

$\rm :\longmapsto\:9{y}^{2} - 9y - 70 = 0$

$\rm :\longmapsto\:9{y}^{2} - 30y + 21y - 70 = 0$

$\rm :\longmapsto\:3y(3y - 10) + 7(3y - 10) = 0$

$\rm :\longmapsto\:(3y + 7)(3y - 10) = 0$

$\rm :\implies\:y = - \dfrac{7}{3} \: \: \: or \: \: \: y = \dfrac{10}{3}$

Consider,

$\rm :\longmapsto\:y = - \dfrac{7}{3}$

$\rm :\longmapsto\:x + \dfrac{1}{x} = - \dfrac{7}{3}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{x} = - \dfrac{7}{3}$

$\rm :\longmapsto\: {3x}^{2} + 3 = - 7x$

$\rm :\longmapsto\: {3x}^{2} + 7x+ 3 = 0$

Its a quadratic equation in x, so to get the values of x, we use Quadratic Formula, given by

$\boxed{ \bf{ \: x = \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}}}$

So, here

$\rm :\longmapsto\:a = 3$

$\rm :\longmapsto\:b = 7$

$\rm :\longmapsto\:c = 3$

So, on substituting the values, we get

$\rm :\longmapsto\:x = \dfrac{ - 7 \: \pm \: \sqrt{ {7}^{2} - 4(3)(3)} }{2(3)}$

$\rm :\longmapsto\:x = \dfrac{ - 7 \: \pm \: \sqrt{ 49 - 36} }{6}$

$\bf :\longmapsto\:x = \dfrac{ - 7 \: \pm \: \sqrt{ 13} }{6}$

Now, Consider

$\rm :\longmapsto\:y = \dfrac{10}{3}$

$\rm :\longmapsto\:x + \dfrac{1}{x} = \dfrac{10}{3}$

$\rm :\longmapsto\: \dfrac{ {x}^{2} + 1}{x} = \dfrac{10}{3}$

$\rm :\longmapsto\: {3x}^{2} + 3 = 10x$

$\rm :\longmapsto\: {3x}^{2} - 10x + 3 = 0$

$\rm :\longmapsto\: {3x}^{2} - 9x - x + 3 = 0$

$\rm :\longmapsto\:3x(x - 3) - 1(x - 3) = 0$

$\rm :\longmapsto\:(x - 3)(3x - 1) = 0$

$\rm :\implies\:x = 3 \: \: \: or \: \: \: x = \dfrac{1}{3}$

Hence,

Solution of

$\rm :\longmapsto\:9\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) - 9\bigg(x + \dfrac{1}{x} \bigg) - 52 = 0 - - (1)$

is

$\bf :\longmapsto\: \boxed{ \bf{ \: x = \dfrac{ - 7 \: \pm \: \sqrt{ 13} }{6}, \: 3, \: \dfrac{1}{3} }}$