Inequalities containing terms in modulus function

Question

Solve the following inequality :- $3x+2<|2x+5|<8x+9$

I need help in verifying if my method of solving is correct

---

I splitted this into 2 inequalities

A) $3x+2<|2x+5|$

Then squared both sides, $(3x+2)^2<(2x+5)^2$

this gave me x in the range $(-7/3 , 3)$

B) similarly did for $|2x+5|<8x+9$ and got x in the range $(-7/5,-2/3)$

The common range comes out to be $(-2/3 , 3)$

Is this process correct to solve inequalities by squaring the terms to remove that modulus function ?

Also, is there any further check I need to put in at the end to ensure that the range is correct if I square the terms initially which could potentially reduce my range?

Answer 1

Doing the exact same thing as you but with domain restrictions will give you the right answer as following :

Case-1 $$|2x+5|<8x+9$$ clearly for this to be true in any case whatsoever $8x+9>0$ must be true. Now squaring under this condition we get

$$(2x+5)^2<(8x+9)^2 \implies(6x+4)(10x+14)>0$$ $$\implies x \in (-\infty,-1.4)\ \cup\ \left(-\frac{2}{3}, \infty \right)$$

Now taking intersection with the $8x+9>0$ condition we get

$$x \in\left(-\frac{2}{3}, \infty \right) $$

---

Case-2 $$3x+2<|2x+5|$$ As noted by @Anne Bauval in the comments one obvious solution in this case would be $3x+2 <0 \implies x < -\frac{2}{3}$ and the other part of solution set comes from squaring the equation as follows $$(3x+2)^2<(2x+5)^2\implies (x-3)(5x+7)<0 \implies x \in \left(-\frac{7}{5} , 3\right)$$

so complete solution set for Case-2 will be the union of both solutions :

$$x \in \left(-\infty , 3\right)$$

---

Taking the intersection of both cases we finally get

$$x \in \left(-\frac{2}{3},3\right)$$

In conclusion always write the domain restrictions and check for obvious cases (applicable to me too ;) ) before squaring up inequalities as squaring always gives botched solutions.

Answer 2

As commented, "squaring both sides" like you did is not correct. Anyway, in this context, introducing and solving quadratic inequations was overcomplicated. Instead, you could apply the three following simple rules: $$|t|=\max(t,-t)$$ $$a<\max(b,c)\iff(a<b\text{ or }a<c)$$ $$\max(b,c)<d\iff(b<d\text{ and }c<d).$$

A) $$\begin{align}3x+2<|2x+5|&\iff(3x+2<2x+5\text{ or }3x+2<-2x-5)\\&\iff x<\max(3,-7/5)=3. \end{align}$$

B) $$\begin{align}|2x+5|<8x+9&\iff(2x+5<8x+9\text{ and }-2x-5<8x+9)\\&\iff x>\max(-2/3,-7/5)=-2/3. \end{align}$$

"The common range comes out to be $(-2/3 , 3)$" indeed,

but this was not the case with your $(-7/3,3)\cap(-7/5,-2/3)=(-7/5,-2/3).$

In the post you mention in comment, "squaring both sides" was legit to solve $|a|\le b$ (because $b$ was known to be positive) but again, overcomplicated compared to solving directly $(a\le b$ and $-a\le b).$ See @AndréNicolas' comment to the non-accepted answer, which was however not quite correct.

Answer 3

You could just use the definition of the modulus: $|2x+5| = 2x+5$ if $x\ge-5/2$ and $|2x+5| = -2x-5$ if $x \lt -5/2$. If $x\ge -5/2$ then the original inequality can be rewritten without the modulus and solved as follows:

$$3x+2\lt 2x+5 \lt 8x+9$$ $$-x+3\lt 5x+7$$ $$ 0\lt 6x+4$$ $$ x\gt -2/3$$

Now to solve for the upper bound:

$$3x+2\lt 2x+5 \lt 8x+9$$ $$-5x-7 \lt -6x-4$$ $$x-3 \lt 0$$ $$x \lt 3$$

Therefore, this inequality holds when $-2/3 \lt x \lt 3$. You can use the definition of the modulus to check the case of $2x+5 \lt 0$ to confirm the result.