3
$\begingroup$

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 $g\left(x\right)=g\left(3\right) = 3$
  • $f\left(x\right)=g\left(x\right) \implies g\left(x\right) \ge 12 \implies x \in \left(-\infty;-6\right)\cup\left(6;+\infty\right)$.
$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ – Nebo Alex Jan 7 '16 at 16:25
  • $\begingroup$ @Boris I don't see how I can do that. Could you elaborate, please? $\endgroup$ – Pavel Vergeev Jan 7 '16 at 16:31
  • $\begingroup$ plot both sides and adjust the value of $a$ so that they intersect at a point $\endgroup$ – Nebo Alex Jan 7 '16 at 16:36
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.