# Find all such $a$ that $x+2\lvert x-3 \rvert = 7\lvert x-a \rvert + 3 \lvert x-a-4|$ has at least one root.

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 $g\left(x\right)=g\left(3\right) = 3$
• $f\left(x\right)=g\left(x\right) \implies g\left(x\right) \ge 12 \implies x \in \left(-\infty;-6\right)\cup\left(6;+\infty\right)$.
• you can solve both sides independently and get a solution .because both sides are straight lines and you must adjust the values of $a$ to find that these lines intersect at a point. – Nebo Alex Jan 7 '16 at 16:25
• @Boris I don't see how I can do that. Could you elaborate, please? – Pavel Vergeev Jan 7 '16 at 16:31
• plot both sides and adjust the value of $a$ so that they intersect at a point – Nebo Alex Jan 7 '16 at 16:36

For big enough $$x$$, the RHS will outgrow the LHS, regardless of the value of $$a$$. This is because the RHS is asymptotic to $$10x$$, while the LHS is asymptotic to $$3x$$.
Furthermore, both sides of our equation are piecewise linear and continuous. Therefore, our question is equivalent to finding the $$a$$’s, such that for some $$x$$, the RHS does not exceed the LHS, and we need only to look at their values at $$x=3,a,a+4$$ to determine them:
• At $$x=3$$, the LHS equals $$3$$, while the RHS equals $$7|3-a|+3|-a-1|$$. It can be verified that no $$a$$ satisfy the desired inequality in this case.
• At $$x=a$$, the LHS equals $$a+2|a-3|$$, while the RHS equals $$12$$. In this case, $$a\in\mathbb{R}\setminus(-6,6)$$ will satisfy the desired inequality.
• At $$x=a+4$$, the LHS equals $$a+4+2|a+1|$$, while the RHS equals $$28$$. It can be verified that the $$a$$ that satisfy the desired inequality in this case are all contained in the solution set of the last case.
Our final answer is therefore that for precisely $$\boxed{a\in\mathbb{R}\setminus(-6,6)}$$ does our original equation have at least a real root.