# Square both sides of an in inequality

I have a simple inequality that looks like:

$$|x+1|\leq|2x+3|$$

I am trying to solve for this by squaring both sides and finding my x values, how come I can't do this? The rule is that if the inequality holds true for all values of x I can square it, but when I solve for my x values this isn't the correct answer. Is this a valid way to solve this type of inequality?

The answer should be $x\leq-2$ or $x \geq -4/3$ but squaring both sides won't give this result.

• Squaring both sides is perfectly valid as both sides are positive. That answer is wrong, it should be $x\leq -2$ or $x\geq -4/3$. E.g. Clearly $x=-1$ satisfies the inequality. – videlity Feb 12 '18 at 0:55
• Yes sorry, forgot the -ve on the 4/3 – user8876747 Feb 12 '18 at 1:01
• I just squared both sides and got the answer. You may have made a mistake in solving it. – videlity Feb 12 '18 at 1:03

Why not? $(|x+1|)^2\le (|2x+3|)^2$, $(x+1)^2\le (2x+3)^2$, $(x+1)^2-(2x+3)^2\le 0$ $$(x+1)^2-(2x+3)^2=(x+1-2x-3)(x+1+2x+3)=(-x-2)(3x+4) \le 0$$ $$x\le-2, x\ge -4/3$$

$a \leq b \Longrightarrow a.c < b.d$ for any $0< c < d$, then $|a| \leq |b| \Longrightarrow |a||a| \leq |b||b|$

So, $|x+1|^2 = (x + 1)^2$ and $|2x + 3|^2 = (2x + 3)^2$. Therefore

$$x^2 + 2x + 1 \leq 4x^2 + 12x + 9 \Longrightarrow 0 \leq 3x^2 + 10x + 8$$

Find the roots. Since $3>0$, if $r_{1}, r_{2}$ are roots, for all $x \in \mathbb{R}\setminus (r_{1}, r_{2})$, $x$ is a solution.

$|a|^2 = a^2$ and $|a| \ge 0$ so of course you can square both sides.

What you can't do is if $x < y$ you can't say that $x^2 < y^2$ because you do not know if $x$ or $y$ are positive or not. If $x < y < 0$ then $x^2 > y^2$. And if $x < 0 <y$ then maybe $x^2 > y^2$ (if $|x| > |y|$ or $x^2 < y^2$ (if $|x| < |y|)$ or $x^2 = y^2$ (if $|x| = |y|$).

$0 \le |x+1| \le |2x + 3|$ so

so yes, $|x+1|^2 = (x+1)^2 \le |2x + 3|^2 = (2x + 3)^2$.

But that might or might not make things easier.

$x^2 + 2x + 1 \le 4x^2 + 12x + 9$

$0 \le 3x^2 + 10x + 8$

To solve, $3x^2 + 10x + 8 =0\implies x = \frac {-10 \pm {100-4*8*3}}6 = \frac {-5\pm\sqrt{25-24}}3 = \frac {-5\pm 1}3 = -2, -\frac 43$

So $(3x^2 -10x + 8) = 3(x+2)(x+\frac 43)$ and

$0 \le (x+2)(x+\frac 43)$.

So either $x + 2 \ge 0$ and $x+\frac 43 \ge 0$ so $x \ge -2$ and $x\ge -\frac 43$, which is redundant so $x \ge - \frac 43$.

Or $x + 2\le 0$ and $x + \frac 43 \le 0$ so $x \le -2$ and $x \le -\frac 43$ so $x \le -2$.

Which is the result you wanted.