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I'm looking for an algebraic solution for $x$.

$$ \frac{x}{x+2} -3 = \frac{5x}{x^2-4}+x$$

My first go at this involved converting this into an expression with a cubic numerator and $(x+2)(x-2)$ as the denominator. To find the roots, I then tried to divide each factor in the denominator into to cubic. No success.

I've converted this expression to:

$$ x(x-7)= (x+3)(x+2)(x-2)$$

which illustrates the futility of my first approach. There are no common factors. Can I solve this without invoking the cubic formula?

Edit: To clarify, the the Precalculus textbook calls for an algebraic and graphic solution. If you have an algebraic solution that would be accessible to a precalculus student, please provide it.

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    $\begingroup$ $x$ is not an integer, so it is much harder to do this without the cubic formula. $\endgroup$ – Landuros Jan 11 '18 at 3:29
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    $\begingroup$ Setting $x=\frac 13(y-\frac 5y-2)$ gives $y^6-362y^3-125=0$ which is a quadratic equation on $y^3$ though I don't know how to get this idea without knowing the exact form of the only real solution. $\endgroup$ – mathlove Jan 11 '18 at 4:24
  • $\begingroup$ If you absolutely must solve it without calculus and cubic formulas, there is always the brute force trial and error solution, applying a table of x values until your converted expression converges to equivalency. Not pure algebra but effective nonetheless. $\endgroup$ – AsymLabs Jan 11 '18 at 11:43
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Assuming $x^2\neq 4$, you end with the cubic equation $$x^3+2 x^2+3 x-12=0$$ So, consider the function $$f(x)=x^3+2 x^2+3 x-12 \implies f'(x)=3 x^2+4 x+3$$ The first derivative does not cancel which means that there is only one real root.

Now, use inspection : $f(0)=-12$, $f(1)=-6$, $f(2)=10$. So, the root is between $1$ and $2$. Looking deeper $f(\frac 32)=\frac 38$ telling that the root is slightly below $1.5$.

Make $x=y+\frac 32$ which makes the equation to be $$g(y)=y^3+\frac{13 y^2}{2}+\frac{63 y}{4}+\frac{3}{8}$$ if we admit that $y$ is small, then $$g(y) \approx \frac{63 y}{4}+\frac{3}{8}=0 \implies y=-\frac{1}{42}$$ So, an approximate solution is $$x \approx \frac 32-\frac{1}{42}=\frac{31}{21}\implies f(\frac{31}{21})=\frac{34}{9261}$$ which is now quite small.

Repeat the process making now $x=y+\frac{31}{21}$ giving $$g(y)=y^3+\frac{45 y^2}{7}+\frac{2270 y}{147}+\frac{34}{9261}$$ then $$g(y) \approx \frac{2270 y}{147}+\frac{34}{9261}=0 \implies y=-\frac{17}{71505}$$ So, an approximate solution is $$x \approx \frac{31}{21}-\frac{17}{71505}=\frac{105538}{71505}\approx 1.47595$$ while the exact solution (solving the cubic) would be $\approx 1.47595$ !

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    $\begingroup$ This is the solution given in the text. Your approach is interesting. I would assume that precalculus students aren't expected to take a derivative. What do you mean when you say it doesn't cancel? $\endgroup$ – Adam Hrankowski Jan 11 '18 at 4:57
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    $\begingroup$ @AdamHrankowski. $f'(x)=0$ is a quadratic equation with no real root. So, in the ral domain, $f(x)$ is just an increasing function with no extremum. So, only one real root. $\endgroup$ – Claude Leibovici Jan 11 '18 at 5:08
  • $\begingroup$ I see. Very nice. Thanks. $\endgroup$ – Adam Hrankowski Jan 11 '18 at 5:17
  • $\begingroup$ I'm not familiar with this process. What's it called? $\endgroup$ – Adam Hrankowski Jan 11 '18 at 5:46
  • $\begingroup$ @AdamHrankowski. This is "almost" the basis of Newton method $\endgroup$ – Claude Leibovici Jan 11 '18 at 6:13
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The posted equation reduces to the cubic $x^3 + 2 x^2 + 3 x - 12 = 0$ which has the real root:

$$ x = \dfrac{1}{3} \left(-2 - \dfrac{5}{\sqrt[3]{181 + 9 \sqrt{406}}} + \sqrt[3]{181 + 9 \sqrt{406}} \right) $$

I do not see how that could be derived without using the cubic formula, or otherwise manually duplicating the steps which lead to it.

If I were to surmise, however, I'd guess that a couple of signs might have got reversed in the transcription. Indeed, the following similar equation reduces to the cubic $x^3- 4 x^2- x + 12 = 0\,$, which has the easy to find rational root $x=3$ and then factors into $\,(x - 3) (x^2 - x - 4) = 0\,$:

$$\frac{x}{x \color{red}{\,\mathbf{-}\,} 2} \color{red}{\,\mathbf{+}\,} 3 = \frac{5x}{x^2-4}+x$$

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  • $\begingroup$ @dxiv I'm tempted by your response. I've double-checked the problem in the text and I've represented it correctly. $\endgroup$ – Adam Hrankowski Jan 11 '18 at 4:44
  • $\begingroup$ @AdamHrankowski Texts can be wrong sometimes, too ;-) Other than that, I really don't see a way to get the closed-form roots of the posted equation, except essentially reinventing the cubic formula. $\endgroup$ – dxiv Jan 11 '18 at 4:49
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    $\begingroup$ Wish the downvoter had left a comment why. $\endgroup$ – dxiv Jan 11 '18 at 4:49
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    $\begingroup$ @AdamHrankowski Thanks, but... This is the solution given in the text Then I suggest you make it more clear in the OP that numerical approximations are acceptable. The way it's written now "looking for an algebraic solution", it reads as if you are asking for exact closed-form solutions. $\endgroup$ – dxiv Jan 11 '18 at 5:04
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    $\begingroup$ @AdamHrankowski a graphing calculator would have been acceptable That probably frames the expectations better. May be worth noting that Claude Leibovici's first iteration gave $x \simeq 31/21 = 1.47619\ldots$ which is $1.48$ to two decimals already. P.S. No worry, didn't think for a second that it was your downvote, and I tried to carefully separate my comment about drive-by downvotes, . $\endgroup$ – dxiv Jan 11 '18 at 5:23

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