Solve 1/(x(x + 2)) − 1/(x + 1)² = 1/12.

Step-by-step solution

1. 1

   Spot the hidden relationship between the two denominators. Expanding, $x(x+2)=x^{2}+2x$, while $(x+1)^{2}=x^{2}+2x+1$. The two differ by exactly $1$, so with

   $u=(x+1)^{2}$

   we have $x(x+2)=u-1$, and the equation collapses to a single-variable problem. Clearing denominators directly would produce a quartic instead.

2. 2

   Rewrite in terms of u.

   $\frac{1}{u-1}-\frac{1}{u}=\frac{1}{12}$

3. 3

   Combine the left side. The numerators telescope:

   $\frac{u-(u-1)}{u(u-1)}=\frac{1}{u^{2}-u}$

   so the equation is $\dfrac{1}{u^{2}-u}=\dfrac{1}{12}$, i.e.

   $u^{2}-u-12=0$

4. 4

   Solve the quadratic in u.

   $(u-4)(u+3)=0\quad\Longrightarrow\quad u=4\ \text{or}\ u=-3$

5. 5

   Reject the impossible value and unwind the substitution. Since $u=(x+1)^{2}$ is a square, $u=-3$ is impossible. From $u=4$:

   $(x+1)^{2}=4\quad\Longrightarrow\quad x+1=\pm 2\quad\Longrightarrow\quad x=1\ \text{or}\ x=-3$

   $\boxed{x=1\ \text{or}\ x=-3}$

6. 6

   Check both roots in the original equation and confirm the domain. The excluded values are $x=0$, $x=-2$ and $x=-1$; neither root is among them. At $x=1$: $\tfrac{1}{1\cdot 3}-\tfrac{1}{4}=\tfrac13-\tfrac14=\tfrac{1}{12}$ ✓. At $x=-3$: $\tfrac{1}{(-3)(-1)}-\tfrac{1}{4}=\tfrac13-\tfrac14=\tfrac{1}{12}$ ✓. The symmetry of the two roots about $x=-1$ is expected, since the whole equation depends on $x$ only through $(x+1)^{2}$.