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

This solution is based on a geometrical reasoning (see picture below)

Let $$\tag{1}f(x):=\dfrac{9x}{x+9} \ \ \text{and} \ \ V_x=\binom{x}{f(x)}.$$

The initial condition can be written under the equivalent form:

$$\tag{2}\|V_x\|^2=40.$$

Said otherwise, we are looking for points on hyperbola $(H)$ given by (1) that are at distance $d:=2\sqrt{10}$ from the origin. Graphical representation indicates that there are two solutions (the abscissas of the intersection points with $(H)$ of the circle centered at the origin with radius $d$).

In order to find them, it is advantageous to rotate the axes by a $\pi/4$ angle around the origin (in this way the abscissas of the intersection points are opposite one to the other ; we can thus expect a degree diminution...)

As this transformation is given by change of coordinates :

$$\tag{3}\begin{cases}x=\tfrac{X-Y}{\sqrt{2}}\\y=\tfrac{X+Y}{\sqrt{2}}\end{cases}$$

(old coordinates expressed as function of new coordinates, as usual), we get the equation of $(H)$ with respect to the new (green) axes, under the form:

$$\tag{4}(Y-9\sqrt{2})^2-X^2=162 \ \ \iff \ \ Y=9\sqrt{2}-\sqrt{162+x^2}$$

(the last equation is the equation of the lower right branch of $(H)$, the only interesting one).

A parameterization of $(H)$ with respect to these new axes is : $$\tag{5}\begin{cases}(a) &X=\tfrac{-18\sqrt{2}t}{t^2-1}\\(b) &Y=tX=\tfrac{-18\sqrt{2}t^2}{t^2-1}\end{cases}$$

Remark : this parameterization is obtained in the classical way (non trivial intersections of $(H)$ with a rotating straight line $Y=tX$).

Expressing (see relationship (2)) that $X^2+Y^2=40$ (rotations preserve norms), we get a fourth degree equation :

$$\tag{6}81(t^2+t^4)=5(t^2-1)^2$$

(with only even powers of $t$, as expected). Setting $T=t^2$, we have a quadratic equation whose only positive solution is $T=\tfrac{1}{19}$ ; thus $t=\pm \tfrac{1}{\sqrt{19}}$. Plugging these values into (5)(a) and (b), we get values for $X$ and $Y$, and finally values of $x$ by using (3):

$$\tag{7} x=1\pm\sqrt{19}$$

which can be checked to be the real roots of the initial fourth degree equation :

$$\tag{8}x^4+18x^3+122x^2-720x-3240=0$$

Remark : Polynomial $x^2-2x-18$ whose roots are given by (7) divides the polynomial in the LHS of (8), with quotient $x^2+20x+180$ whose roots $x=-10 \pm i 4 \sqrt{5}$ are the complex roots of (8).

Answer 5

Two days of thinking:

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

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

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

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

UPD: fixed a typo

Answer 6

Hint:

If $\dfrac{9x}{x+9}=y,$

$xy=?, x- y=?$

Now use $x^2+y^2=(x-y)^2+2xy$