# Difference between the equation inequalities and absolute value inequalities

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

• Welcome to Maths SX! I don't understand the rules which are applied to remove tha absolute values. Can you explain? (using MathJax please!) – Bernard Feb 9 at 15:48
• @Bernard i have basically put into two equation one is x+4<= 2x+10 and the other to x+4<= -(2x+10) – Dev Patel Feb 9 at 15:50
• But you have to argue according to the sign of $x+4$! – Bernard Feb 9 at 15:52
• @bernard i didnt get you sorry – Dev Patel Feb 9 at 15:53
• 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). – 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?

• i had 0<= 3x^2 + 32x +84 – Dev Patel Feb 9 at 16:03
• Not in your link. If you had this, it shows the problem is reduced to a quadratic inequation, which is standard from high school. – Bernard Feb 9 at 16:06
• i was thinking of doing it to – Dev Patel Feb 9 at 16:07
• If you do it, you'll find the correct result. – Bernard Feb 9 at 16:12
• i got this 0<= -4 2/3 and 0 <= -6 – 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 $$-x-4\le 2(-x-5)$$

• i am not getting the consideration – Dev Patel Feb 9 at 15:46
• 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 – Dev Patel Feb 9 at 15:58
• And the other cases? – Dr. Sonnhard Graubner Feb 9 at 16:03
• do you mean the LHS or another x – Dev Patel Feb 9 at 16:05
• Not another $x$ only an interval. – 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.

• i am wandering how to know when to use the normal case or the case using quadratic eqaution – Dev Patel Feb 9 at 16:59
• Unless you know the terms are positive, you dont transform it into a quadratic inequality. – Sabrosky Feb 9 at 17:26
• 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. – Robert Howard Feb 9 at 17:41
• @Sabrosky if i have -2x i have to use quadratics and of i have +2x i can use the normal way. – Dev Patel Feb 10 at 3:32