Consider the following question:

What is the sum of all possible solutions of the equation $|x + 4|^2 - 10|x + 4| = 24$?

The answer is $-8$. I was able get $-8$ by doing it the regular way - checking $10|x + 4|$ for positive and negative values. However, when I follow the critical points/transitional points approach I'm getting the answer as $2$ which is clearly wrong. What am I doing wrong here?

I was learning about critical point approach from this link.

These were my steps using the critical point approach (which definitely does not appear to be correct):
- critical points x <= -4 and x >= 4
- when x <= -4 you get x = -6 or x = 8 after factoring out the equation. Since x has to be <= -4 only -6 is valid in this approach.
- When x >= 4 the only valid value would x = 8. So -6 + 8 = 2

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    $\begingroup$ What's the critical points/transitional points approach? $\endgroup$ – tilper Apr 9 '16 at 22:30
  • $\begingroup$ Please see my edit. There was another website that called it transitional point approach. Not sure which one is right. If you think it should be called something else please let me know so I can edit it. $\endgroup$ – user3587180 Apr 9 '16 at 22:34
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    $\begingroup$ This technique appears to be for solving inequalities. Can you show or explain the steps you do to obtain the incorrect answer of 2? $\endgroup$ – tilper Apr 9 '16 at 22:48
  • $\begingroup$ What you are saying could be true (the technique is used for in-equalities). I'm not a math major so I have no idea about this stuff :) $\endgroup$ – user3587180 Apr 9 '16 at 23:02
  • $\begingroup$ Where do 4 and -4 come from? Based on that link, the so-called critical points should be values of $x$ that satisfy the equation. Neither 4 nor -4 satisfy it. And even when you do find the values of $x$ that satisfy the equation, you just need to check for and throw out extraneous solutions and then add up the actual solutions. No need to consider inequalities at all. $\endgroup$ – tilper Apr 9 '16 at 23:30

The technique described in the link does not apply to this problem because the technique is about solving inequalities. This problem requires no use of inequalities.

I also don't really understand the technique you mentioned where you checked $10|x+4|$ for positive and negative values. I know what that means but I don't know how it helps.

The best way (that I can think of) to approach this is to simply solve the equation and add up the valid solutions.

First let $y=|x+4|$ and subtract 24 from both sides. Then the equation becomes $y^2 - 10y -24=0.$ Solve by factoring (or however) to get $y=-2$ and $y=12.$

Back-substitute to solve for $x$. The two equations you'll get are $|x+4|=-2$ and $|x+4|=12.$ The first one has no solutions because the LHS is an absolute value and can therefore never be equal to a negative number. The second one yields solutions $x=8$ and $x=-16.$ Verify both solutions in the original equation from the beginning of the problem to make sure we don't have extraneous solutions. We don't. Therefore the solutions are 8 and -16, and so the sum is -8.

Some background info on critical points, etc. You asked about what this method should be called. As with a few things in precalculus, there's really no standard name for this type of technique. I've heard critical points, critical values, critical numbers, key values, and now transitional points. Different books and instructors say different things but AFAIK, high-level research rarely (if ever) requires in-depth discussion of topics like this, which may be why there's no standard name. Personally I find it unfortunate that "critical points" (or "critical" anything) is used in this context because the terms critical point and critical value actually do have a standard definition in calculus, and it's very different from what's happening here in this problem. This calculus definition is what Moo was getting at with his comment to your original question but, as I stated in my follow-up comment, the fact that it gets the correct answer is a coincidence here.

  • $\begingroup$ Thanks for the answer! I don't agree that it is coincidence though. This question came up when I was preparing for GMAT test. I have seen other similar questions solved using this approach. Look at "3-steps approach for complex problems" from this link. Maybe that would give you more information. gmatclub.com/forum/math-absolute-value-modulus-86462.html $\endgroup$ – user3587180 Apr 10 '16 at 2:04
  • $\begingroup$ No problem. I assure you it's a coincidence. Those 3 steps aren't related to what Moo was getting at, because they still aren't related to calculus critical points. Calculus critical points are (roughly) the zeros of the derivative of a function. In general, the zeros of the derivative have nothing to do with the zeros of the function itself. I see now why you were thinking of inequalities but I feel like those 3 steps overcomplicate the process of solving absolute value equalities. If $|x| =a$ where $a$ is not negative, then $x=a$ or $x=-a.$ Same principle applies (to be continued) $\endgroup$ – tilper Apr 10 '16 at 2:10
  • $\begingroup$ (continued due to length) even if we have something more complicated than just $x$ inside the absolute value signs. $\endgroup$ – tilper Apr 10 '16 at 2:10

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