# Help on this inequation

#### Utherr

$$\displaystyle \sqrt{-x-2}-\sqrt{x+5}<3$$ 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 $$\displaystyle x\in(-\infty,-2]$$

#### Prove It

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

#### 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

#### HallsofIvy

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

#### Prove It

MHF Helper
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.

• HallsofIvy

$$\displaystyle \sqrt{-x-2}-\sqrt{x+5}<3$$ 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 $$\displaystyle x\in(-\infty,-2]$$
If you want to use the 6th roots

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

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

$$\displaystyle x=-2\ \Rightarrow\ \sqrt{0}-\sqrt{3}<3$$
$$\displaystyle x=-3\ \Rightarrow\ \sqrt{1}-\sqrt{2}<3$$
$$\displaystyle x=-4\ \Rightarrow\ \sqrt{2}-\sqrt{1}<3$$
$$\displaystyle x=-5\ \Rightarrow\ \sqrt{3}-\sqrt{0}<3$$
$$\displaystyle x=-6\ \Rightarrow\ \sqrt{4}-\sqrt{-1}=3$$