# 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.

• Many users here do not like "being told" what to do (for example: "Solve...." or "Prove...". Your questions will be more warmly received if you actually frame a question as a question, and if you include what you've tried to do before posting. – amWhy Nov 8 '12 at 21:13
• You might want to start by multiplying both sides of your equation by $(x+9)^2$. – amWhy Nov 8 '12 at 21:15
• That's an improvement. "Please help me to solve..." would be a next step. Please read this faq. Also, when suggestions are made or someone asks you for clarification, try to interact or respond, so we can provide help, accordingly. – amWhy Nov 8 '12 at 21:22
• The problem text has $81 x^2$ but the title has $9 x^2$ - Please fix. – NoChance Nov 8 '12 at 21:32
• Suggestion (rather than hint): $a^2+b^2=(a+bi)(a-bi)$.40=36+4=4+36. – Mark Bennet Nov 8 '12 at 22:44

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

• Looks promising! I haven't the time to work my way back to find the $x_i$. For now, +1 for effort! I've already spent too much time on this question, etc... – amWhy Nov 9 '12 at 0:24
• Looks right to me, and quite "clean" in getting to a "quadratic of a quadradic" case, simpler than the full blown degree 4 equation. +1 – coffeemath Nov 9 '12 at 0:38
• @ coffeemath,thanks – Adi Dani Nov 9 '12 at 0:40

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?

• I fail to see that insult but sorry I guess. – Jorge Lavín Nov 8 '12 at 21:26
• @Burzum He probably refers to $81+81-40$... While many people who asks questions need to see that, I can see why a strong student would find that insulting ;) – N. S. Nov 8 '12 at 21:41
• I don't see how anyone could find that insulting. If the OP thinks this material is too easy for him/her, they should not be asking here. – wj32 Nov 8 '12 at 21:48
• @amWhy I guess I might had also insulted him/her :) ...wj32 Which material? Reducing this rational equation to a fourth degree polynomial, or figuring out what 81+81-40 is? ;) – N. S. Nov 8 '12 at 21:58
• I think you might have saved several people a lot of time if you had put what is in the above comment in your original question. – Todd Wilcox Nov 8 '12 at 22:36

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

• Checked that already! No rational roots! – 17SI.34SA Nov 8 '12 at 22:37
• didn't intend to be insulting...The thing is, without clarification about what you've tried, and what you already know about the problem, it's hard to know what you need help with, so people start with the basics...in terms of hints/etc. So, for example, the first recommendation was getting the quadratic out of the denominator...then simple algebra, then seeing if one can find integer roots...etc. – amWhy Nov 8 '12 at 22:46
• Hints are made for the comment section! Am I right? Unless, if you sure that your hint will lead to the answer... – 17SI.34SA Nov 8 '12 at 22:50
• re: "hints belong in comments". No, not necessarily. Because, again, without any context or clarification about where you are, when it looks like a problem encountered in an assignment, many prefer to begin with hints. If you made clear from the start what you've done and ruled out, everything would have been a lot smoother. At this point, many, I suspect, have "walked away" from this problem. We really do just want to help...No one intended to insult. – amWhy Nov 8 '12 at 22:54
• I retagged your post, accordingly, given your response to @MarkBennet . – amWhy Nov 8 '12 at 22:58

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).

Hint:

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

$xy=?, x- y=?$

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

• I see that our interest has been triggered by this old issue, and you are also considering an old and a new equation. I have tried in my answer to give a geometrical motivation to the operations to be done. – Jean Marie Jan 5 '18 at 20:47

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

• The only problem is that there is no common factor of $x^2 - 2x - 18$ appearing in the third expression - i.e. $x^2 + 20x + 180 \neq x^2 + 20x - 180$ and multiplying the factors in the final expression gives $= x^4+18x^3-238x+3240$. What you end with has four real roots. What you start with has 2 non-real roots. The two real roots of line 1 = the two real roots of line 4. See link to Wolfram in my answer, if you're interested. – amWhy Nov 9 '12 at 0:02
• @ amWhy,what do you think about my solution – Adi Dani Nov 9 '12 at 0:12
• @AdiDani I'll look... :-) – amWhy Nov 9 '12 at 0:15