Can we solve this without cubic formula?

Question

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.

Answer 1

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 $\approx 1.47595$ !

Answer 2

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 $\,(x - 3) (x^2 - x - 4) = 0\,$:

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