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The problem

Solve the equation: $||x-7|-15|=10$

My idea

I took 2 cases:

Case $1$: $|x-7|-15=10\implies |x-7|=25\implies$

  • (a) $ x-7=25\implies x=32$

  • (b) $ x-7=-25\implies x=-18$

Case $2$: $|x-7|-15=-10\implies |x-7|=5\implies$

  • (a) $x-7=5\implies x=12$

  • (b) $ x-7=-5\implies x=2$

So the solution I got were $x=32,-18,12,2$

I'm not sure if taking the cases like this is a correct reasoning.

PrincessEev
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    Don't put the tag in the title. It's already associated to the post. – Cameron L. Williams Sep 04 '24 at 14:11
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    I think its good. You consider all the possibilities, so there cannot be other solution to the equation. But if you want to be very strict, then you need to verify the answer you found satisfied the original equation, but it is straight forward. – Angae MT Sep 04 '24 at 14:15
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    In MathJax use \implies to get $\implies$ instead of using => – jjagmath Sep 04 '24 at 14:16
  • Let $|x-7|=A\geq0$. Solve for $A$. You did similar tough. – Bob Dobbs Sep 04 '24 at 14:24
  • If you don't like breaking down absolute value equations into cases, a trick you can use is squaring, because $\forall x\in\mathbb{R}: |x|^2 = x^2$. With two absolute value operators, you get a quartic equation, which for this question works out to $x^4 - 28x^3 - 356x^2 + 7728x - 13824 = 0$. Though, this may just be making things more complicated. – Dan Sep 04 '24 at 15:23
  • I'm not sure how one would do it without cases and doing it in cases is certainly correct reasoning. In one line: $||x-7| -15|=10\implies |x-7|-15=\pm 10\implies |x-7|=5,25\implies x-7=\pm 5, \pm 25\implies x = 2, -12, -18,32$. On thing to note though if you got $|x-7|< 0$ you'd have to rule that out (example $||x-7|-15|=17\implies |x-7|-15 =\pm 17\implies |x-7|=-2,32$ but $-2$ is impossible so $|x-7|=32\implies x=\pm 32\implies x=-35, 39$. (But I'm sure you would have caught that.) – fleablood Sep 04 '24 at 16:56
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    @Dan I never understood such strong adversion to cases. Squaring is a way to avoid that be it potentially adds so many possible extraneous solutions and it's just so... indirect it seems like cutting off you nose to spite your face. (Although it is a valid strategy) – fleablood Sep 04 '24 at 16:59
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    Your question is lacking focus. Please edit your post to let us know precisely why you are "not sure if taking the cases like this is a correct reasoning". https://math.stackexchange.com/questions/tagged/solution-verification: For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is too broad or missing context. Instead, the question must identify precisely which step in the proof is in doubt, and why so. – Anne Bauval Sep 04 '24 at 18:10

3 Answers3

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Taking cases like this is perfectly reasonable, especially considering how the absolute value is defined: $$ |x| \stackrel{\text{def.}}{=} \begin{cases} x, & x > 0 \\ -x, & \le 0 \end{cases} $$ You can even see that your work leads to the correct solution by graphing both sides of the equation:

enter image description here

(Link to Desmos demo.)

PrincessEev
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You can avoid cases in this kind of problem by using squarings. But that does not guarantee this is economical.

Squaring the original equation and putting in $|u|^2=u^2$ gives

$(|x-7|-15)^2=100$

Expand the left side and isolate the remaining $|x-7|$ factor:

$(x-7)^2-30|x-7|+225=100$

$30|x-7|=(x-7)^2+125=x^2-14x+174$

Squaring again removes the final absolute value signs and you have a polynomial equation:

$900(x-7)^2=(x^2-14x+174)^2$

$x^4-28x^3-356x^2+7728x-13824=0$

which we can solve for possible roots of $x$ by seeking rational roots. We find the roots given in other answers ($2,12,-18,32$) and they all check out.

And now you know why absolute value problems are more commonly solved via case-by-case analysis.

Oscar Lanzi
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Your way is fine and as an alternative to improve it and also to check the solution we can observe that the LHS function is equal to zero at $x=22$ and at $x=-8$ and it is symmetric with respect to $x=7$ then for $x\ge 7$

$$||x-7|-15|=10 \iff |x-22|=10$$

which has solutions $x= 12$ and $x=32$ and then by symmetry with respect to $x=7$ also $x=2$ and $x=-18$ are solutions.

user
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  • This does not answer the question. The OP asked (admittedly too vaguely) for a feedback on their own method. – Anne Bauval Sep 04 '24 at 18:16
  • @AnneBauval Thanks my idea was give a feedback for an improvement and also to check. I explain it better. Thanks – user Sep 04 '24 at 18:55