# How do you solve \frac { 2x - 1} { x + 2} - \frac { 1} { x } \leq 1?

Dec 23, 2017

$x : \left(- 2 , 2 - \sqrt{6}\right] \cup \left(0 , 2 + \sqrt{6}\right]$

#### Explanation:

$\frac{2 x - 1}{x + 2} - \frac{1}{x} \le 1$

$\frac{x}{x} \cdot \frac{2 x - 1}{x + 2} - \frac{x + 2}{x + 2} \cdot \frac{1}{x} \le 1$

$\frac{\left(2 {x}^{2} - x\right) - \left(x + 2\right)}{x \left(x + 2\right)} \le 1$

$\frac{2 {x}^{2} - 2 x - 2}{x \left(x + 2\right)} \le 1$

At this point, we must look at three cases. One where x is positive, and one where x is between -2 and 0, and one where x is less than -3. This is because the denominator reads $x \left(x + 2\right)$. As you may know, multiplying an inequality by a negative value changes the direction of the inequality so we must take these cases.

We also know that $x \ne 0$ and $x \ne 2$ because then we would be dividing by $0$.

Case 1: Assume $x > 0$. Then $x \left(x + 2\right) > 0$. Thus, no sign changes.

$2 {x}^{2} - 2 x - 2 \le {x}^{2} + 2 x$

${x}^{2} - 4 x - 2 \le 0$

From this equation, we need to determine when it is less than 0. To do so, we can figure out where the quadratic intersects the x-axis. Since it opens up (the coefficient of the ${x}^{2}$ term is greater than 0), we know that the x values between the zeroes are where the quadratic is negative.

$x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \cdot 1 \cdot \left(- 2\right)}}{2 \cdot 1}$

$x = \frac{4 \pm \sqrt{24}}{2}$

$x = 2 \pm \sqrt{6}$

That means the quadratic is negative for x-values on the interval $\left(2 - \sqrt{6}\right) , \left(2 + \sqrt{6}\right)$.

We only take the positive case because $2 - \sqrt{6} < 0$, but we defined $x > 0$ for this case. Thus, we have found one interval:

$x : \left(0 , 2 + \sqrt{6}\right]$

Case 2; $x : \left(- 2 , 0\right)$. If this is true, then $x < 0$ but $x + 2 > 0$. Thus, the term $x \left(x + 2\right)$ is multiplying a negative and a positive, so $x \left(x + 2\right) < 0$. That means we have to flip the direction of the inequality when we multiply by $x \left(x + 2\right)$.

$2 {x}^{2} - 2 x - 2 \ge {x}^{2} + 2 x$

${x}^{2} - 4 x - 2 \ge 0$

Based on the math above, we know the quadratic is positive or zero when $x \le 2 - \sqrt{6}$ or $x \ge 2 + \sqrt{6}$. Since we know $x : \left(- 2 , 0\right)$, we have found another interval.

$x : \left(- 2 , 2 - \sqrt{6}\right]$.

Finally, case 3: $x < - 2$. In this case, both $x$ and $x + 2$ are negative, so their product is positive. That means our algebra and solutions will look very similar to Case 1.

$x : \left(2 - \sqrt{6}\right) , \left(2 + \sqrt{6}\right)$

Since the interval of solutions does not occur in our case of $x < - 2$, there are no solutions in this case.