Solve the following equation: $\frac{1}{x^2}+\frac{1}{(4-\sqrt{3}x)^2}=1$

Solve the following equation: $$\frac{1}{x^2}+\frac{1}{(4-\sqrt{3}x)^2}=1$$

I know it's from a Math Olympiad but I don't know which and I couldn't find it on the internet. Expanding everything doesn't work (it leads you to a 4th degree equation with no rational roots). I don't know what else to try, I can't see any useful substitution.

Any help would be appreciated, thanks.

• I haven't solved it, but does this help?-->:$x^2(3x-2\sqrt3)(x-2\sqrt3)=-7$ – Vikram Jul 29 '14 at 8:12

If you do a substitution $t = x \sqrt 3$, you get $$\frac 3 {t^2} + \frac 1{(4 - t)^2} = 1 \implies t^4 - 8t^3 + 12 t^2 + 24t - 48=0$$ You can check that $t = 2$ is a solution, so $P_4(t) = (t-2)P_3(t)$, therefore $x = \frac 2{\sqrt 3}$ is a solution of the initial equations.

• But a cubic polynomial always has at least one real root. In fact $P_3(t)=3t^3-6t^2+24$ has critical points at $(0,24)$ and $(4,-8)$, and since it is increasing it must have three real roots. If by complex you meant ugly, that sounds about right, WolframAlpha does give nasty solutions... – Deathkamp Drone Jul 29 '14 at 8:52
• Oops, $P_3(t)=t^3-6t^2+24$. – Deathkamp Drone Jul 29 '14 at 8:57
• @DeathkampDrone dang, you're absolutely right. I was too lazy to check the solution Mathematica provided, but after full simplification turned out all roots are real, and moreover - not a bit nasty. Not sure how to solve it though. Using Cardano seems like an overkill to me. Prolly you can use some trig substitution. – Kaster Jul 29 '14 at 17:19
• Not really worth a full answer, but since there are no solutions when $t$ is a large negative or a large positive, and $f(0) < f(1) < 0$, using the rational root theorem leaves a couple of integer solutions, one of which is $t = 2$. – Toby Mak Oct 13 '19 at 6:49

It seems like you're right, Kasper. A trigonometric substituition would solve the equation completely. I'll write the solution I arrived at:

Finding inequalities is helpful to motivate the right subtituition. We know that both $\frac{1}{x^2}$ and $\frac{1}{(4-\sqrt{3}x)^2}$ are positive, so we can say that: $$\frac{1}{x^2}<1\Leftrightarrow \left|\frac{1}{x}\right|<1\Leftrightarrow -1<\frac{1}{x}<1$$ And analogously $\displaystyle-1<\frac{1}{4-\sqrt{3}x}<1$. The substituition is starting to become obvious. One last hint is that we're trying to find two squares whose sum equals $1$. This immediately reminds us of the identity $\sin^2 a+\cos^2 a=1$. That's enough to motivate the substituition (WLOG): $$\frac{1}{x}=\sin a \,\,\text{ and }\,\,\frac{1}{4-\sqrt{3}x}=\cos a$$

Which will give us: \begin{align} &\frac{1}{\displaystyle4-\frac{\sqrt{3}}{\sin a}}=\cos a \Leftrightarrow \\\\ &\frac{1}{\cos a}+\frac{\sqrt{3}}{\sin a} = 4 \Leftrightarrow \\\\ &\frac{1}{2}\sin a + \frac{\sqrt{3}}{2}\cos a = 2\sin a\cos a \Leftrightarrow \\\\ &\sin(60°+a)=\sin(2a) \end{align}

Solving and substituting back, the solutions are: $$x_1=\csc60°=\frac{2}{\sqrt{3}},\;x_2=\csc 40°,\;x_3=\csc20°,\;x_4=-\sec10°.\;\;\blacksquare$$

• Excellent trick with the trig substitution. Nice job. I personally think this one should be accepted as an answer. – Kaster Aug 4 '14 at 0:20
• Thanks. I'm sure how the whole "accepting an answer" works, but I was assuming OP is supposed to accept the one that helped him the most, so I accepted yours. – Deathkamp Drone Aug 4 '14 at 4:26

Let $u = \sqrt{3}x$, and $v = 4 - \sqrt{3}x$, then:

$u + v = 4$, and $\dfrac{3}{u^2} + \dfrac{1}{v^2} = 1 \to 3v^2 + u^2 = u^2v^2 \to 3v^2 = u^2(v^2 - 1) \to 3v^2 = (4-v)^2(v^2 - 1)$. Observe that $v=2$ is a root to the above equation. From this we can use synthetic division to factor the polynomial and finish it.

One solution is "clear" at $\frac{2}{\sqrt{3}}$. I was motivated to look for something like this by trying to write $1$ as the sum of two simple fractions, and the presence of $3$ and $4$.

It's also halfway between the two vertical asymptotes of $\frac1{x^2}+\frac{1}{\left(4-\sqrt{3}x\right)^2}$, and in a sketch of that function (which clearly is never negative) it was natural to want to see what the output was at that midpoint.