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

Question

As in the title,

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

Thanks.

Answer 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 $$x_i=\frac{9}{t_i-1},i=1,2,3,4$$

Answer 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?

Answer 3

Update: Omitted unhelpful hints (ground apparently already covered by OP!)

Edit: given the correspondence below, Perhaps the link below may be of help?

It doesn't show how to solve your problem...it only reveals what those solutions are... two real, to non-real solutions. Don't click on the link unless you want to know the solutions. They are not numerical approximations. Perhaps knowing the solutions will allow you to work "backwards" so to speak, to get the "how."

Wolfram Alpha solutions

Answer 4

Two days of thinking:

$$x^4 + 18x^3 + 122x^2 - 720\cdot x - 3240 =$$

$$= x^4 + (20x^3 - 2x^3)+(180x^2-40x^2-18x^2)-(360x+360x) -3240$$

$$=x^2(x^2+20x-180) - 2x(x^2+20x+180)-18(x^2+20x+180)$$

$$=(x^2+20x-180)(x^2-2x-18)$$