I didn't multiple anything by x.
No. I don’t think you did multiply by x. Rather you multiplied by 4x^2, which entailed multiplying by a positive number. But earlier you squared fractions, which raises the same issue in an even more complex way.
You attacked the problem in an unnecessarily complex way.
You started off well by thinking about the domain issues and got [imath]x \ne 0.[/imath]
Now you could have simplified your life with
[math]1 - 4x^2 \ge 0 \implies 1 \ge 4x^2 \implies \dfrac{1}{4} \ge x^2 \implies[/math]
[math]\sqrt{\dfrac{1}{4}} \ge \sqrt{x^2} \implies \left | \dfrac{1}{2} \right | \ge |x|.[/math]
Putting those two together, you end up with
[math]- \dfrac{1}{2} \le x < 0 \text { or } 0 < x \le \dfrac{1}{2}.[/math]
This just indicates the domain. We still have to deal with the inequality under two different cases. Now multiplying by 2x to clear fractions is the simplest approach.
[math]0 < x \le \dfrac{1}{2} \text { and } \dfrac{1 - \sqrt{1 - 4x^2}}{x} \implies 2 - 2\sqrt{1 - 4x^2} > 3x \implies[/math]
[math]2 - 3x > 2\sqrt{1 - 4x^2} \ge 0 \implies 4 - 12x + 9x^2 > 4 - 16x^2 \implies 25x^2 > 12x[/math]
But x is positive by hypothesis.
[math]25x > 12 \implies \dfrac{12}{25} < x \le \dfrac{1}{2}.[/math]
And then you solve the case where x is negative.
Nothing wrong in how you did the problem. Correct answers do not care how you find them. But you will make fewer errors with a simpler approach.
