Question
Solve $2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)=11$
✓
Answer
$2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)=11$
Let $x+\frac{1}{x}=y$
squaring on both side
$x^2+\frac{1}{x^2}=y^2-2$
Putting these values in the given equation
$2\left(y^2-2\right)-y=11 $
$\Rightarrow2 y^{\wedge} 2-4- y -11=0^`$
$ \Rightarrow 2 y^2-y-15=0$
$ \Rightarrow 2 y^2-6 y+5 y-15=0 $
$ \Rightarrow(y-3)(2 y+5)=0$
If $y - 3 = 0$ or $2y + 5 = 0$
then $y = 3$ or $y = (-5)/2$
$\Rightarrow x+\frac{1}{x}=3 \quad \text { or } x+\frac{1}{x}=\frac{-5}{2}$
$\Rightarrow \frac{x^2+1}{x}=3 \text { or } \frac{x^2+1}{x}=\frac{-5}{2}$
$ \Rightarrow x^2-3 x+1=0 \text { or } 2 x^2+5 x+2=0$
$\Rightarrow x=\frac{-3 \pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}$ or $2 x^2+4 x+x+2=0$
$\Rightarrow x=\frac{-3 \pm \sqrt{5}}{2}$ or $2 x(x+2)+1(x+2)=0$
then $x =-2$ and $x=\frac{-1}{2}$
Let $x+\frac{1}{x}=y$
squaring on both side
$x^2+\frac{1}{x^2}=y^2-2$
Putting these values in the given equation
$2\left(y^2-2\right)-y=11 $
$\Rightarrow2 y^{\wedge} 2-4- y -11=0^`$
$ \Rightarrow 2 y^2-y-15=0$
$ \Rightarrow 2 y^2-6 y+5 y-15=0 $
$ \Rightarrow(y-3)(2 y+5)=0$
If $y - 3 = 0$ or $2y + 5 = 0$
then $y = 3$ or $y = (-5)/2$
$\Rightarrow x+\frac{1}{x}=3 \quad \text { or } x+\frac{1}{x}=\frac{-5}{2}$
$\Rightarrow \frac{x^2+1}{x}=3 \text { or } \frac{x^2+1}{x}=\frac{-5}{2}$
$ \Rightarrow x^2-3 x+1=0 \text { or } 2 x^2+5 x+2=0$
$\Rightarrow x=\frac{-3 \pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}$ or $2 x^2+4 x+x+2=0$
$\Rightarrow x=\frac{-3 \pm \sqrt{5}}{2}$ or $2 x(x+2)+1(x+2)=0$
then $x =-2$ and $x=\frac{-1}{2}$
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