Solve the following inequality

Question

I have the following inequality $\frac {2x^2}{x+2} < x-2$. I tried to solve it the with the following steps.

step 1 $\frac {2x^2}{x+2} < x-2$

step 2 $\frac {2x^2}{x+2} - (x-2) < 0$

step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} < 0$

step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2} < 0$

step 5 $\frac {2x^2 - x^2 + 4}{x+2} < 0$

step 5 $\frac {x^2 + 4}{x+2} < 0$

step 6 I used character study to get the result x > 2. But this is incorrect.

Where did I go wrong with this?. I feel that I made a mistake somewhere in step 2 but not sure what I did wrong.

Thanks!

Answer 1

HINT : $\frac{x^2+4}{x+2}\lt 0$ is correct. Then, note that $x^2+4\gt 0$ for any $x\in\mathbb R$.

Answer 2

$\left(\frac{x^2+4}{ x+2}\right) < 0$ implies $\left(\frac{1}{x+2}\right) < 0$ because $x^2 + 4 >0$ which implies $x+2 < 0$ and then $x < -2$ Please note $\left(\frac{1}{x+2}\right) = 0$ can never occur.

Answer 3

When you arrive at $$\frac{f(x)}{g(x)}<0$$ multiply $f(x)$ and $g(x)$ and create a table of values. Make sure $g(x) \neq 0$ though.

Answer 4

I would like to suggest another way: you can define two cases, one where $x>-2$, the other where $x<-2$, and multiply the initial inequality by $(x+2)$. It is then straightforward to get the final result.

Answer 5

Your inequality holds good if $x>-2$, not $2$.