# Thread: Help on this inequation

1. ## Help on this inequation

$\sqrt{-x-2}-\sqrt[3]{x+5}<3$ In this inequation what i'd usually do is bring -x-2 and x+5 to the "same root" that's 6. But i don't think i'm going anywhere because of that 3 on the right. What would you suggest me to do here?

the condition is $x\in(-\infty,-2]$

2. Have you tried graphing the functions $y = \sqrt{-x-2} - \sqrt[3]{x + 5}$ and $y = 3$ and determining the $x$ values for which this inequality holds true?

3. i need to prepare for an exam, i don't think doing graphs will help me if i have to solve a similar inequation

4. Solve the equation $\sqrt{-2-x}+ \sqrt[3]{x+ 5}= 3$. The points where those are equal or where the roots are not defined (x must be less than -2, of course) separate ">" from "<".

5. Originally Posted by Utherr
i need to prepare for an exam, i don't think doing graphs will help me if i have to solve a similar inequation
On the contrary, you should NEVER try to solve any equations or inequations without picturing what you are actually dealing with (i.e. graphing). To gain a deep understanding of function analysis, you need to understand and be able to connect the numerical, graphical and algebraic representations of your function.

6. Originally Posted by Utherr
$\sqrt{-x-2}-\sqrt[3]{x+5}<3$ In this inequation what i'd usually do is bring -x-2 and x+5 to the "same root" that's 6. But i don't think i'm going anywhere because of that 3 on the right. What would you suggest me to do here?

The condititon is $x\in(-\infty,-2]$
If you want to use the 6th roots

$\sqrt{-x-2}=(-x-2)^{\frac{3}{6}}=\left(\sqrt[6]{-x-2}\right)^3$

$\sqrt[3]{x+5}=(x+5)^{\frac{2}{6}}=\left(\sqrt[6]{x+5}\right)^2$

$x=-2\ \Rightarrow\ \sqrt{0}-\sqrt[3]{3}<3$

$x=-3\ \Rightarrow\ \sqrt{1}-\sqrt[3]{2}<3$

$x=-4\ \Rightarrow\ \sqrt{2}-\sqrt[3]{1}<3$

$x=-5\ \Rightarrow\ \sqrt{3}-\sqrt[3]{0}<3$

$x=-6\ \Rightarrow\ \sqrt{4}-\sqrt[3]{-1}=3$