In the equation, $a$ is a parameter and $x$ is a variable: $$x+2\lvert x-3 \rvert = 7\lvert x-a \rvert + 3 \lvert x-a-4|.$$ I want to find all values of $a$ that make the equation have at least one real root.
Context: My textbook says this can be accomplished by looking at the restrictions of the functions on both sides of the equation. The only thing I can think of is to find their min/max values.
What I've done: I found some restrictions but failed to come up with a solution:
- $f\left(x\right) = 7\lvert x-a \rvert + 3 \lvert x-a-4|$; min $f\left(x\right)=f\left(a\right)=12$
- $g\left(x\right) = x+2\lvert x-3 \rvert$; min $g\left(x\right)=g\left(3\right) = 3$
- $f\left(x\right)=g\left(x\right) \implies g\left(x\right) \ge 12 \implies x \in \left(-\infty;-6\right)\cup\left(6;+\infty\right)$.