# How do i go about solving this? |2x−1| > √(1−x^2)

## I have tried spiting to 2x−1 > √(1−x^2) and 2x−1 < -√(1−x^2) and solving for x, but the second part does not work out, or at least i cant figure it out

Jan 20, 2018

$- 1 \le x < 0$ or $\frac{4}{5} < x \le 1$

#### Explanation:

$\sqrt{1 - {x}^{2}}$ has domain $- 1 \le x \le 1$.

$| 2 x - 1 |$ can be written as a piecewise function:

$2 x - 1$ when $x \ge \frac{1}{2}$ and $1 - 2 x$ when $x < \frac{1}{2}$ so we're solving two equations:

$1 - 2 x > \sqrt{1 - {x}^{2}}$ on the interval $- 1 \le x < \frac{1}{2}$

and

$2 x - 1 > \sqrt{1 - {x}^{2}}$ on the interval $\frac{1}{2} \le x < 1$

For the first inequality, let's first solve the equation:

$1 - 2 x = \sqrt{1 - {x}^{2}}$

Square both sides:

$1 - 4 x + 4 {x}^{2} = 1 - {x}^{2}$

move everything to one side:

$5 {x}^{2} - 4 x = 0 \rightarrow x \left(5 x - 4\right) = 0$

So $x = 0$ and $x = \frac{4}{5}$ are the intersection points.

Since the inequality is only valid on $- 1 \le x < \frac{1}{2}$, $x = 0$ is the only one that matters.

So consider the intervals $- 1 \le x < 0$ and $0 < x < \frac{1}{2}$. Testing points in those regions tells us that $1 - 2 x > \sqrt{1 - {x}^{2}}$ on the first interval and $1 - 2 x < \sqrt{1 - {x}^{2}}$ on the second interval.

So far our solution set is $- 1 \le x < 0$.

Now let's look at the second inequality.

First we'll solve the equality for the intersection points:

$2 x - 1 = \sqrt{1 - {x}^{2}}$

square both sides:

$4 {x}^{2} - 4 x - 1 = 1 - {x}^{2}$

moving everything to one side:

$5 {x}^{2} - 4 x = 0 \rightarrow x \left(5 x - 4\right) = 0$

so the intersections are $x = 0$ and $x = \frac{4}{5}$.

Since this inequality is only valid on $- \frac{1}{2} \le x \le 1$, we only want $x = \frac{4}{5}$.

So consider the intervals $\frac{1}{2} \le x < \frac{4}{5}$ and $\frac{4}{5} < x < 1$. Testing points in those regions tells us that $2 x - 1 < \sqrt{1 - {x}^{2}}$ on the first interval and $2 x - 1 > \sqrt{1 - {x}^{2}}$ on the second interval. Since we want the greater than part, we add $\frac{4}{5} < x \le 1$ to our solution set.

Overall the solution set is $- 1 \le x < 0$ or $\frac{4}{5} < x \le 1$.