Help on this inequation

Jun 2010
44
5
\(\displaystyle \sqrt{-x-2}-\sqrt[3]{x+5}<3\) 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 \(\displaystyle x\in(-\infty,-2]\)
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Have you tried graphing the functions \(\displaystyle y = \sqrt{-x-2} - \sqrt[3]{x + 5}\) and \(\displaystyle y = 3\) and determining the \(\displaystyle x\) values for which this inequality holds true?
 
Jun 2010
44
5
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
Apr 2005
20,249
7,909
Solve the equation \(\displaystyle \sqrt{-2-x}+ \sqrt[3]{x+ 5}= 3\). The points where those are equal or where the roots are not defined (x must be less than -2, of course) separate ">" from "<".
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
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.
 
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Dec 2009
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\(\displaystyle \sqrt{-x-2}-\sqrt[3]{x+5}<3\) 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 \(\displaystyle x\in(-\infty,-2]\)
If you want to use the 6th roots

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

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

Instead...

\(\displaystyle x=-2\ \Rightarrow\ \sqrt{0}-\sqrt[3]{3}<3\)

\(\displaystyle x=-3\ \Rightarrow\ \sqrt{1}-\sqrt[3]{2}<3\)

\(\displaystyle x=-4\ \Rightarrow\ \sqrt{2}-\sqrt[3]{1}<3\)

\(\displaystyle x=-5\ \Rightarrow\ \sqrt{3}-\sqrt[3]{0}<3\)

\(\displaystyle x=-6\ \Rightarrow\ \sqrt{4}-\sqrt[3]{-1}=3\)