Solve the equation: 6 (x^2+1 x^2 )-25 (x-1 x )+12=0. - Vidyadip

Question

Solve the equation:
$6\left(x^2+\frac{1}{x^2}\right)-25\left(x-\frac{1}{x}\right)+12=0$.

✓

Answer

Given equation
$6\left(x^2+\frac{1}{x^2}\right)-25\left(x-\frac{1}{x}\right)+12=0$
Put $x -\frac{1}{x}=y$, squaring $\left(x-\frac{1}{x}\right)^2=y^2$
$\Rightarrow x^2+\frac{1}{x^2}-2=y^2 $
$ \Rightarrow x^2+\frac{1}{x^2}=y^2+2$
Now, given equation becomes
$6\left(y^2+2\right)-25 y+12=0 $
$ \Rightarrow 6 y^2+12-25+12=0 $
$ \Rightarrow 6 y^2-25 y+24=0 $
$ \Rightarrow 6 y^2-16 y-9 y+24=0$
$ \Rightarrow 2 y(3 y-8)-3(3 y-8)=0 $
$ \Rightarrow(3 y-8)(2 y-3)=0 $
$\Rightarrow 3 y-8=0 \text { or } 2 y-3=0$
$ \Rightarrow 3 y=8 \text { or } 2 y=3 $
$ \Rightarrow y=\frac{8}{3} \text { or } y=\frac{3}{2}$
But $x-\frac{1}{x}=y$
$\therefore x-\frac{1}{x}=\frac{8}{3} $
$\Rightarrow \frac{x^2-1}{x}=\frac{8}{3} $
$\Rightarrow 3 x^2-3=8 x $
$\Rightarrow 3 x^2-8 x -3=0 $
$\Rightarrow 3 x ^2-9 x + x -3=0 $
$\Rightarrow 3 x ( x -3)+1( x -3)=0 $
$\Rightarrow( x -3)(3 x +1)=0 $
$\Rightarrow x -3=0 \text { or } 3 x +1=0 $
$\Rightarrow x =3 \text { or } x =\frac{-1}{3}$
or
$x-\frac{1}{x}=\frac{3}{2} $
$ \Rightarrow \frac{x^2-1}{x}=\frac{3}{2}$
$ \Rightarrow 2 x^2-2=3 x $
$ \Rightarrow 2 x^2-3 x-2=0$
$ \Rightarrow 2 x^2-4 x+x-2=0 $
$\Rightarrow 2 x(x-2)+1(x-2)=0$
$ \Rightarrow(x-2)(2 x+1)=0 $
$\Rightarrow x-2=0 \text { or } 2 x+1=0 $
$ \Rightarrow x=2 \text { or } x=\frac{-1}{2} $
$\text { Hence, } x=3, \frac{-1}{3}, 2 \text { and } \frac{-1}{2} .$

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