How one should solve $x^2+\frac{81x^2}{(x+9)^2}=40$ [closed]

As in the title,

Please help me solve $x^2+\frac{81x^2}{(x+9)^2}=40$

Thanks.

Solution 1:

Equation $x^2+\frac{81x^2}{(x+9)^2}=40$ can be written in form $$x^2+\left(\frac{9}{1+\frac{9}{x}}\right)^2=40$$

if we replace $1+\frac{9}{x}=t$ then we have that $$\tag{1} x=\frac{9}{t-1}$$ our equation becomes $$\frac{1}{(1-t)^2}+\frac{1}{t^2}=\frac{40}{9^2}$$ denote $$\tag{2} A=\frac{40}{9^2}$$ multiplying both sides by $t^2(t-1)^2$ after rearranging we get $$2(t^2-t)+1=A(t^2-t)^2$$ then we use that

$$\tag{3} y=t^2-t$$ or

$$ Ay^2-2y-1=0$$ the solutions are$$y_1=\frac{1+\sqrt{A+1}}{A};y_2=\frac{1-\sqrt{A+1}}{A}$$ from (3) we get following equations $$t^2-t-y_1=0$$ with roots $t_1,t_2$ $$t^2-t-y_2=0$$ with roots $t_3,t_4$ finally from (1) we get $$x_i=\frac{9}{t_i-1},i=1,2,3,4$$

Solution 2:

Multiply by $(x+9)^2$ both sides

$x^2(x+9)^2+81x^2=40(x+9)^2$

$x^2(x^2+81+18x)+81x^2-40(x^2+81+18x)=0$

$x^4+18x^3+x^2(81+81-40)-40·18x-40·81=0$

Can you continue from here?