Using symbol lab I put this in $|x+4|\le |2x+10|$ and the answer I get is $x \le -6$ or $x\ge -14/3$, but when I manually worked out it was $\;x \ge -6\;$ or $\;x\ge -14/3$. My working out is in the description:The image of my working out

  • $\begingroup$ Welcome to Maths SX! I don't understand the rules which are applied to remove tha absolute values. Can you explain? (using MathJax please!) $\endgroup$ – Bernard Feb 9 at 15:48
  • $\begingroup$ @Bernard i have basically put into two equation one is x+4<= 2x+10 and the other to x+4<= -(2x+10) $\endgroup$ – Dev Patel Feb 9 at 15:50
  • $\begingroup$ But you have to argue according to the sign of $x+4$! $\endgroup$ – Bernard Feb 9 at 15:52
  • $\begingroup$ @bernard i didnt get you sorry $\endgroup$ – Dev Patel Feb 9 at 15:53
  • $\begingroup$ You didn't take into account that; for instance, $|x+4|=x+4$ or $-x-4$, depending on the values of $x$. Similarly for $|2x+10|$. So your second line of computation is not equivalent to the first (the given inequation). $\endgroup$ – Bernard Feb 9 at 16:02

The simplest way to solve this inequation is to use the fact that, for any $A$, $|A|^2=A^2$ (this removes the absolute values), and that function $x^2$ is increasing for non-negative $x$. Thus $$|x+4|\le |2x+10|\iff(x+4)^2\le (2x+10)^2\iff3x^2+32x+84\ge 0.$$ Can you proceed?

  • $\begingroup$ i had 0<= 3x^2 + 32x +84 $\endgroup$ – Dev Patel Feb 9 at 16:03
  • $\begingroup$ Not in your link. If you had this, it shows the problem is reduced to a quadratic inequation, which is standard from high school. $\endgroup$ – Bernard Feb 9 at 16:06
  • $\begingroup$ i was thinking of doing it to $\endgroup$ – Dev Patel Feb 9 at 16:07
  • $\begingroup$ If you do it, you'll find the correct result. $\endgroup$ – Bernard Feb 9 at 16:12
  • $\begingroup$ i got this 0<= -4 2/3 and 0 <= -6 $\endgroup$ – Dev Patel Feb 9 at 16:12

You must consider the following cases: $$x\geq -4$$ and $$x\geq -5$$ or $$x\geq -4$$ and $$x<-5$$ or$$x<-4$$ and $$x<-5$$ If $$x\geq -4$$ and $$x\geq -5$$ then we have to solve $$x+4\le 2(x+5)$$ Can you proceed? The second case is not possible so you have to solve for $$-5\le x<-4$$: $$-x-4\le 2(x+5)$$ and for $$x<-5$$ you have to solve $$-x-4\le 2(-x-5)$$

  • $\begingroup$ i am not getting the consideration $\endgroup$ – Dev Patel Feb 9 at 15:46
  • $\begingroup$ what i see is you have factorised the RHS but my steps will be the same and will result in 2x+10, which is the same to what i did $\endgroup$ – Dev Patel Feb 9 at 15:58
  • $\begingroup$ And the other cases? $\endgroup$ – Dr. Sonnhard Graubner Feb 9 at 16:03
  • $\begingroup$ do you mean the LHS or another x $\endgroup$ – Dev Patel Feb 9 at 16:05
  • $\begingroup$ Not another $x$ only an interval. $\endgroup$ – Dr. Sonnhard Graubner Feb 9 at 16:07

Here you have a simple difference:

|𝑥+4|≤|2𝑥+10| <=> (𝑥+4)^2 ≤ (2𝑥+10)^2, as both sides are positive.

On the other hand, if 𝑥+4 ≤ 2𝑥+10 it is not necessarily true that it is equivalent to (𝑥+4)^2 ≤ (2𝑥+10)^2, as x+4 and 2x+10 can have any sign.

For example, -2 ≤ 1 is not equivalent to 4 ≤ 1.

This is the reason that you get different answers, as you are changing the condition you are giving the set of numbers that satisfy the inequality.

  • $\begingroup$ i am wandering how to know when to use the normal case or the case using quadratic eqaution $\endgroup$ – Dev Patel Feb 9 at 16:59
  • $\begingroup$ Unless you know the terms are positive, you dont transform it into a quadratic inequality. $\endgroup$ – Sabrosky Feb 9 at 17:26
  • $\begingroup$ Welcome to MSE, Sabrosky! Could you please edit your answer to include MathJax formatting? It's standard practice here, and it makes your answer both nicer-looking and easier to read. $\endgroup$ – Robert Howard Feb 9 at 17:41
  • $\begingroup$ @Sabrosky if i have -2x i have to use quadratics and of i have +2x i can use the normal way. $\endgroup$ – Dev Patel Feb 10 at 3:32

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