Solve 2x + 2(lambda)x + mu = 0 for x, treating lambda and mu as parameters.

Step-by-step solution

1. 1

   Collect the $x$ terms and factor. Both variable terms carry a factor $2x$:

   $2x+2\lambda x=2x(1+\lambda)\quad\Longrightarrow\quad 2x(1+\lambda)+\mu=0$

2. 2

   Isolate the product. Subtract $\mu$:

   $2x(1+\lambda)=-\mu$

3. 3

   Divide only when the coefficient is nonzero. Dividing by $2(1+\lambda)$ requires $1+\lambda\neq 0$:

   $x=\frac{-\mu}{2(1+\lambda)},\qquad\lambda\neq-1$

   Dividing without checking is the error that hides the whole second half of the answer.

4. 4

   Handle the degenerate case $\lambda=-1$. Then the coefficient of $x$ is zero and $x$ disappears entirely:

   $2x-2x+\mu=0\quad\Longrightarrow\quad\mu=0$

   So if $\mu=0$ the equation is $0=0$ and every real $x$ is a solution; if $\mu\neq 0$ it is a contradiction and there is no solution.

5. 5

   Check the generic formula numerically. Take $\lambda=2$, $\mu=-1$: then $x=\tfrac{1}{6}$, and $2\left(\tfrac16\right)+4\left(\tfrac16\right)-1=\tfrac13+\tfrac23-1=0$ $\checkmark$. Take $\lambda=0$, $\mu=3$: $x=-\tfrac32$, and $-3+0+3=0$ $\checkmark$.