Solve (x − 2x/(x + 2))² + 4x²/(x + 2) = 5.

Step-by-step solution

1. 1

   Simplify the bracket first — it is the whole trick. Putting the two terms over the common denominator $x+2$:

   $x-\frac{2x}{x+2}=\frac{x(x+2)-2x}{x+2}=\frac{x^{2}+2x-2x}{x+2}=\frac{x^{2}}{x+2}$

   So the messy bracket is just $\dfrac{x^{2}}{x+2}$.

2. 2

   Introduce the substitution. Let

   $y=\frac{x^{2}}{x+2}$

   Then the second term is $\dfrac{4x^{2}}{x+2}=4y$, and the equation becomes a plain quadratic:

   $y^{2}+4y=5\quad\Longrightarrow\quad y^{2}+4y-5=0$

3. 3

   Solve for y.

   $(y+5)(y-1)=0\quad\Longrightarrow\quad y=1\ \text{or}\ y=-5$

4. 4

   Unwind y = 1.

   $\frac{x^{2}}{x+2}=1\ \Longrightarrow\ x^{2}=x+2\ \Longrightarrow\ x^{2}-x-2=0\ \Longrightarrow\ (x-2)(x+1)=0$

   giving $x=2$ or $x=-1$ (both allowed, since neither is $-2$).

5. 5

   Check whether y = −5 gives real solutions.

   $\frac{x^{2}}{x+2}=-5\ \Longrightarrow\ x^{2}=-5x-10\ \Longrightarrow\ x^{2}+5x+10=0$

   Its discriminant is $25-40=-15<0$, so this branch contributes no real roots. Discarding it without checking would be the one place an answer could go wrong.

   $\boxed{x=2\ \text{or}\ x=-1}$

6. 6

   Verify in the original equation. At $x=2$: the bracket is $2-\tfrac{4}{4}=1$ and the second term is $\tfrac{16}{4}=4$, so $1+4=5$ ✓. At $x=-1$: the bracket is $-1-\tfrac{-2}{1}=1$ and the second term is $\tfrac{4}{1}=4$, so $1+4=5$ ✓.