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problem

Solve the equation $|4x-3|+ |4x+3|=6$

my idea

First we wrote $|4x-3|+ |4x+3|= |-4x+3|+ |4x+3|$ and used the triangle inequality ($|a|+|b| \geq |a+ b|$ with equality only when both a and b are positive or negative at the same time) and we obtained that $|4x-3|+ |4x+3|= |-4x+3|+ |4x +3| \geq 6$ with equality when $-4x+3, 4x+3 \geq 0$ which means that $x \in [\frac{-3}{4}, \frac{3}{4}]$ or when $-4x+3, 4x+3 \leq 0$ which was no solution.

I'm not sure if I did correctly when I applied the case of equality. Should I take both cases when $\geq 0$ and when $<0$?

Brian Tung
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    Compare with similar computations, e.g., here, or here, or here, or here. – Dietrich Burde Sep 04 '24 at 14:53
  • There are three cases to deal with: $x \le -\frac{3}{4}$, $-\frac{3}{4} \le x \le \frac{3}{4}$, and $x \ge \frac{3}{4}$. – Dan Sep 04 '24 at 14:57
  • @DietrichBurde Thanks. But i see none of the sources use my idea, I'm interested in knowing if my idea is correct... –  Sep 04 '24 at 14:58
  • By variying the signs, there are potentially four cases to discuss, but only three of them are possible (because $4x-3<4x+3$ is guaranteed). –  Sep 04 '24 at 15:04
  • As it is a strict equality, using an inequality will not be precise enough and add a lot more cases. Just do it. There are four cases $|4x-3| = \pm (4x-3)$ and $|4x+3|=6$ so just solve $\pm (4x-3) + (\pm (4x+3)) = 6$. (then check for those that $\pm(4x-3)$ and $\pm(4x+3)$ are non negative. – fleablood Sep 04 '24 at 15:11
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    @fleablood Yeah, but the inequality also says that the cases of equality occurs when both have the same sign...doesn't it make a correct solution? –  Sep 04 '24 at 15:12
  • OP: Hi! I removed the modules tag because as far as I can tell, it's not relevant. Did you read the tag description? – Brian Tung Sep 04 '24 at 15:28
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    Your solution looks valid to me, and I like it. If this were a problem in a precalculus course, I'd expect solutions more like the answers below. – Matthew Leingang Sep 04 '24 at 15:42
  • Your solution is original and smart indeed. Where precisely is your doubt? Is it about strict or large inequality in the second case? – Anne Bauval Sep 04 '24 at 17:54
  • You have posted a very similar question at the same time. Take time to breath... – Jean Marie Sep 04 '24 at 18:05
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    @JeanMarie These two questions are not "very similar". In each, the OP asked for a feedback on their method, which is really different. The method in the linked post is much more standard. – Anne Bauval Sep 04 '24 at 18:43

2 Answers2

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This function is piecewise linear and its graph is a polyline (unbounded). We can solve the problem by looking for crossings between the horizontal $y=6$ and the line segments (or half-lines) that constitute the polyline. The vertices are found where the arguments of the absolute values are zero.

Notice that $4x-3=0$ implies $4x+3=6$ and symmetrically. The vertices are: $$(-\infty,\infty), (-\tfrac34,6), (\tfrac34,6), (\infty,\infty).$$

We conclude that the solution is the whole interval $[-\tfrac34,\tfrac34])$.

enter image description here

0

Despite my doubts your way does work.

If $(-4x+3)$ and $(4x+3)$ are the same sign we get

$|(-4x+3)+(4x+3)| = |(-4x+3)|+|4x+3| = 6$ and as $|(-4x+3) + (4x+3)| = |6|=6$ this is certain always true if $-4x+3$ and $4x+3$ are the same sign.

If $(-4x+3)$ and $(4x+3)$ are opposite signs we get $|(-4x+3)+(4x+3)|> |-4x+3| + |4x+3| = 6$ but as $|(-4x+3) + (4x+3)| = |6|=6$ this can never be true.

So it is true if either $-4x + 3\ge 0$ and $4x+3 \ge 0$ or if $-4x+3 < 0$ and $4x+3 < 0$. There former is $x\in [-\frac 34,\frac 34]$ and the latter is contradictory. Hence your solution.

But in my opinion this is overkill.

For one thing in real numbers the triangle inequality is can be much more strongly stated that as $|A+B| = \begin{cases} \max(|A|,|B|)-\min(|A|,|B|)&\text{which is }\le |A|+|B|\\|A|+|B|\end{cases}$.

and it'd be easier to do it in cases and

This is a case the $x$s "cancel out". I don't think another case, such as $|4x+3| + |3x-4| = 7$ would work as nice.

(Let's see. Either $|4x+3 +(-3x+4)|=|x+7|=7$ and $-\frac 34 \le x \le \frac 43$ and $x=\pm 7$. Impossible. so $|4x+3+(-3x+4)|=|x+7|> 7$ while $4x+3$ and $-3x+4$ or opposite signs. In this case we have an inequality that will not give a precise answer.)

fleablood
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