Inequality $x^2\leq y$

Question

I have a question about how to handle this inequality: $$x^2\leqslant y \to x\leqslant \pm\sqrt{y}$$ or it should be $$\sqrt{x^2}\leq \sqrt{y}\Rightarrow$$ either $$x\leq\sqrt{y}$$ or $$-x\leq\sqrt{y}\Rightarrow x\geq-\sqrt{y}$$ so$$-\sqrt{y}\leq x\leq\sqrt{y}$$ Is my way of thinking is correct?

Answer 1

If $y<0$ $$ x^2 \le y$$ does not have any real solution.

For $y=0$, the only solution is $x=0$

For $ y>0$, we get $$ x^2 \le y \iff |x|\le \sqrt y \iff -\sqrt y \le x \le \sqrt y $$

Answer 2

I assume you work in $\mathbb{R}$. If $y$ is negative, then no $x$ satisfies the inequality. If $y$ is non-negative, then $-\sqrt{y} \leq x \leq \sqrt{y}$.

Answer 3

$$x^2\le y\to x^2-y\le0$$ $$\to (x+\sqrt y)(x-\sqrt y)\le0$$ Thus critical values are $x=\sqrt y, x=-\sqrt y$

Assuming $y\in\Bbb R, y\ge0$,test $x=\sqrt y+1, x=0, x=-\sqrt y-1$

$$x=\sqrt y+1\to(\sqrt y+1)^2=y+2\sqrt y+1>y \text{ (fail)}$$ $$x=0\to 0^2\le y \text{ (success)}$$ $$x=-\sqrt y-1\to(-\sqrt y-1)^2=[-(\sqrt y+1)^2]=y+2\sqrt y+1>y \text{ (fail)}$$

So: $$-\sqrt y \le x \le \sqrt y$$