So I was bored and tackled a bonus exercise during math class, and I managed to derive that the solution could be found as the solution for $x$ in the following equation: $$\frac{\sqrt{x^2-9}+3\sqrt{x^2-1}}{x^2}-\frac12\sqrt{3}=0$$ Using my Graphing calculator, I find that $x\approx4.16331999$. The square of that is approximately $17.33333333$, so I suspect that the value of $x$ should be $\sqrt{17\frac13}$. I could substitute the $x$ in the equation above to verify this, but I'm not satisfied with that.

Instead, I was wondering if there was a way to solve the equation for $x$ instead of just making a guess using a calculator. However, I don't know how to solve the equation since because I can't seem to get rid of the square roots.

I am aware that there might be another way to solve the original problem, but it bothers me that I don't know how to solve the equation above even though an algebraic solution seems to exist.

How would I solve an equation like the one above without using a calculator?


For those interested, the original problem was the following:

An equilateral triangular flag is hung between two poles, one with height 4 and one with height 3. Two corners of the flag are attached to tops of the poles, and the remaining corner touches the ground. What is the length of the sides of the flag?

It was the last and hardest question in the "Wiskunde Olympiade" (Math Olympics) of 2007, and 0% of the participants answered correctly.

  • $\begingroup$ Put $x^2=t$, could be helpful in reducing the degree. $\endgroup$ – Shubham May 27 '14 at 11:56
  • $\begingroup$ Was the original problem to solve an $8^{th}$ degree equation in x, or maybe $4^{th}$ degree in $y$, with $y=x^2$ ? $\endgroup$ – tpb261 May 27 '14 at 11:58
  • $\begingroup$ You can eliminate roots by rearranging (square roots on the left), squaring, rearranging again (the only remaining square root on the left) and squaring again. Each squaring increasing the degree of equation and introduces new "fake" solutions. Solve the equation and test with the original equation each of the candidates to see which is the correct one. Of course what @shubham suggests should help remove unnecessary additional complexity. $\endgroup$ – orion May 27 '14 at 12:29

The equation you want to develop being $$\frac{\sqrt{x^2-9}+3\sqrt{x^2-1}}{x^2}-\frac12\sqrt{3}=0$$ start setting $x^2=t$ as suggested by Shubham; so it becomes $$\frac{\sqrt{t-9}+3\sqrt{t-1}}{t}-\frac12\sqrt{3}=0$$ Multiply everything by $t$ and move the second part to the rhs. The expression becomes $$\sqrt{t-9}+3\sqrt{t-1}=\frac {\sqrt{3}}{2}t$$ Let us define $a=\sqrt{t-9}$,$b=3\sqrt{t-1}$,$c=\frac {\sqrt{3}}{2}t$. So we have now $$a+b=c$$ Square both sides and develop, moving $a^2$ and $b^2$ to the rhs. So $$2ab=c^2-a^2-b^2$$ and square again for reaching finally $$4a^2b^2=(c^2-a^2-b^2)^2$$ and now replace. Before simplification, the result is $$-\frac{9 t^4}{16}+15 t^3-91 t^2=0$$ $t^2$ can be factored and you end with $$(-\frac{9 t^2}{16}+15 t-91) t^2=0$$ The roots of the quadratic equation are $t=\frac{28}{3}$ and $t=\frac{52}{3}$. We must discard $t=0$ and $t=\frac{28}{3}$ because of the domain definition. So, the only root left is $t=\frac{52}{3}$ and, since $t=x^2$, the solutions are $$x_{\pm}=\pm \sqrt \frac{52}{3}$$ Just be patient.

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  • $\begingroup$ I understand that $t=0$ can be discarded because the original equation involved division by $t$, but could you elaborate why $t=\frac{28}{3}$ can be discarded as well (I'm not sure what you mean by the domain definition)? I checked and I noticed that the original equation isn't true for $t=\frac{28}{3}$, so why does this come up as a potential solution? $\endgroup$ – RPFeltz May 28 '14 at 7:34
  • $\begingroup$ Because ot the square roots, $t>9$; that's all you need. $\endgroup$ – Claude Leibovici May 28 '14 at 7:45
  • $\begingroup$ I know that $t$ should be greater than $9$, but $\frac{28}{3}=9\frac13$, which is greater than $9$. $\endgroup$ – RPFeltz May 28 '14 at 9:24
  • $\begingroup$ Oooops ! Sorry for this enormous mistake about $28/3$ ! In fact, when you start squaring radicals, you may introduce spurious roots (just as orion commented). This is the case in your problem. Thanks for pointing my stupidity ! Cheers. $\endgroup$ – Claude Leibovici May 28 '14 at 9:28
  • $\begingroup$ Ah, I see... So the reason there are false roots is because you already knew whether the root was positive or negative before you squared it, and squaring the root obscures that information? Then, is it possible to take note of which root should be used later on, or does that get too complicated? $\endgroup$ – RPFeltz May 28 '14 at 9:33

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