# Question about a solution to a problem involving $x^2\lt a$

If $$a \gt 0$$ show that the solution set of the inequality $$x^2 \lt a$$ consists of all numbers $$x$$ for which $$-\sqrt a \lt x \lt \sqrt a$$. Solution: $$x^2 \lt a \Longrightarrow$$ $$x^2 - a \lt 0\Longrightarrow$$ $$(x + \sqrt a)(x - \sqrt a) \lt 0$$

If $$x \lt -\sqrt a$$ then $$x + \sqrt a \lt 0$$ and $$x -\sqrt a \lt -2\sqrt a \lt 0$$ therefore $$(x + \sqrt a)(x - \sqrt a) \gt 0$$ which is a contradiction. There is more to the solution but my question is why is $$-2\sqrt a$$ used? Is the solution valid with just $$x -\sqrt a \lt 0$$ instead of $$x -\sqrt a \lt -2\sqrt a \lt 0$$? It would seem to yield the same contradiction. Thanks

• $<0=$ is not a good notation. Sep 27, 2022 at 16:12
• @DietrichBurde what should be used? Sep 27, 2022 at 16:31
• Better is $\Longrightarrow$ or $\Longleftrightarrow$. Sep 27, 2022 at 16:51

There is $$-2\sqrt{a}$$ because $$x>-\sqrt{a}$$ (the assumption) and $$x$$ is estimated upward to $$-\sqrt{a}$$. The upwards estimated difference is hence $$-2\sqrt{a}$$. Both are practically the same, $$-2\sqrt{a}$$ is one reasoning step.