Square Root Inequality with param

Question

I don't understand, what should I do to solve?

Do I need to make a replacement in the solution?

Answer 1

First you have to suppose $a^2-x^2\ge0$ and $2x\ge0$ so \begin{equation} 0 \le x\le \lvert a \rvert \end{equation} After you can put the square root at the right side $$2x>\sqrt{a^2-x^2}$$ that it means: $$4x^2>a^2-x^2.$$ This is equivalent to $$x^2>\frac{a^2}{5}.$$ That it means $$x<-\frac{\lvert a \rvert}{\sqrt 5} \quad \mbox{and} \quad x>\frac{\lvert a \rvert}{\sqrt 5}$$ Now, considering the before statement we have:

$$\frac{\lvert a \rvert}{\sqrt 5} < x \le \lvert a \rvert $$

Answer 2

$$0\le\sqrt{a^2-x^2}<2x\to \\a^2-x^2<4x^2\to 5x^2>a^2\to\\x>\dfrac{|a|}{\sqrt 5}\\or\\x<-\dfrac{|a|}{\sqrt 5}$$since because of $0\le\sqrt{a^2-x^2}<2x$ we have $x>0$ then $x<-\dfrac{|a|}{\sqrt 5}$ is invalid therefore:$$\dfrac{|a|}{\sqrt 5}<x\le |a|$$

Answer 3

If $a=0$ we have $2x>\sqrt{-x^2},$ which has no solutions.

Let $a\neq0$.

Thus, the domain gives $$a^2-x^2\geq0$$ or $$-|a|\leq x\leq|a|.$$

Since $$2x>\sqrt{a^2-x^2}\geq0,$$ we obtain $x>0$, which gives $$0<x\leq|a|.$$

Now, squaring gives $$4x^2>a^2-x^2$$ or $$x^2>\frac{a^2}{5},$$ which with domain gives $$\frac{|a|}{\sqrt5}<x\leq|a|.$$

We got the answer:

If $a=0$ then $\varnothing$;

If $a\neq0$ then $\left(\frac{|a|}{\sqrt5},|a|\right]$.