Solving following irrational inequaltiy Here is the inequality: $$x-3<(x^2+4x-5)^{1/2}.$$
My try: First of all I know for RHS to be defined $x^2+4x-5$ should be greater than $0$,so I got the intervals $[-\infty,-5]\cup[1,\infty]$, then I squared both sides and got $x>(7/5)$. So the overall answer is $((7/5),\infty)$ but this is not correct. So my question is where am I wrong and how should I correct?
 A: I would probably think like this. If $x-3<0$ and $x^2+4x+5\geq 0$ then the inequality certainly holds. This is $x<3$ and (if you analyse the geometry of the quadratic) $x\leq -5$ or $x\geq 1$. 
This means that the inequality certainly holds for $x\leq -5$ and $1\leq x<3$.
If you can't follow this you should draw a schematic graph and/or numberline.
Now what if $x-3\geq 0\Rightarrow x\geq 3$?  Now assume that $x$ satisfies the above inequality $I_1$ and $x\geq 3$. Well in this case both sides are positive and squaring is order preserving so if you square both sides you get a new inequality $I_2$ (such that all solutions of $I_1$ satisfy $I_2$). This new inequality has the solution $x\geq 7/5$. So we have 
$$\underbrace{[x\geq 3]}_{=:A}\text{ and }\underbrace{[x\text{ satisfies }I_1]}_{=:B}\Rightarrow [x\text{ satisfies }I_2]\Rightarrow [x\geq \frac75]\Rightarrow \underbrace{[x\geq 3]}_{=A} \text{ or }\underbrace{[\frac75\leq x<3]}_{=:C}.$$
So you have a situation where $A$ and $B$ implies something stronger than $A$, namely $A$ or $C$ ($A\cap B\Rightarrow A\cup C$). Therefore $A\Rightarrow B$ ... somebody might be able to explain this logic a little better than I... it can be understood using truth tables or Venn diagrams. 
EDIT: Henning has explained that this particular logic is flawed.
Therefore the answer is $\mathbb{R}\backslash (-5,1)=(-\infty,-5]\cup[1,\infty)$.
So when your root is defined --- outside $(-5,1)$ --- your root is positive and you have three situations:


*

*for $x\leq -5$, $x-3$ is negative and so less than any root

*for $1\leq x <3 $, $x-3$ is negative and so less than any root

*for $x\geq 3$, $x-3$ is positive but less than this particular square root


Outside these regions the root is undefined so there are no solutions.
A: When you square both sides of the inequality, you can only be sure to preserve the ordering if both sides were non-negative in the first place.
The square root is always nonnegative if it exists, of course, but $x-3$ is only nonnegative for $x\ge 3$. So you need to divide into two cases:


*

*When $x<3$, then $x-3$ is negative, and therefore automatically smaller than the square root whenever the square root exists.

*When $x\ge 3$, then your square-and-rearrange is valid, so in that case you need the square root to exist and $x>\frac75$. But both of these conditions are true for every $x\ge 3$.
So no matter whether $x\ge 3$ or not, the inequality will be satisfied whenever the square root exists.
