# what is the concept used in solving irrational inequalities?

for example to solve for x, $$\frac{\sqrt{2x-1}}{x-2}<1$$

what are the basic steps I need to remember while solving such questions?

• Some things to keep in mind: $\sqrt{A}$ only makes sense if $A\geq 0$ and the left hand side only makes sense unless $x\not = 2$. It might be useful to split into cases; $x>2$ and $x<2$ and analyze them seperately. – Winther Sep 6 '16 at 17:35

The domain of the equation is defined by the conditions: $$\;x\ge \frac12$$ and $$x\neq 2$$.

We can multiply both sides by $$(x-2)^2$$, which is positive on the domain of validity of the inequation: $$\frac{\sqrt{2x-1}}{x-2}<1\iff \sqrt{2x-1}< x-2$$ Thus we have two cases:

1. $$\frac12\le x<2$$: the left-hand side (f the original inequation) is negative, hence the inequality is satisfied.
2. $$x>2$$: both sides are positive, so \begin{align} \sqrt{2x-1}< x-2 &\iff 2x-1 < (x-2)^2\iff x^2-6x+5>0\\&\iff(x-1)(x-5)> 0\iff\begin{cases}x<1\\\enspace\text{or}\\x>5\end{cases} \end{align} We retain only the solutions $$x>5$$

Summing up these results, we obtain the following set of solutions: $$\bigl[\tfrac12,2\bigr)\cup(5,+\infty)$$.

• I stumbled upon this question and your very clear answer. I just have an additional question regarding case (1). I see your line of reasoning, but say I want to proceed mechanically. Thus, should I move as follows? First get $\sqrt{2x-1}>x-2$, since $x-2<0$ and then solve. But then I get as a solution $1 < x < 5$ and if I put it in the system, I get that (1) is satisfied by $[1/2 , 1)$, which is wrong. Where is the problem? I just don't see it. Thanks a lot for any feedback! :) – Kolmin Sep 11 '17 at 18:10
• @Kolmin: The problem is in squaring. To compare two numbers, you can do it by squaring only if both numbers have the same sign, which isn't the case. Just take an example: we have $1 >-2$ and $1^2<(-2)^2=4$. We also have $1>-1/2$ and $1^2>(-1/2)^2=1/4$. – Bernard Sep 11 '17 at 18:28
• Oh Right. Thanks a lot! But then, in a sense, there is no way to get a mechanical solution for this problem. You must see that in (1) the LHS is negative hence smaller than $1$. Or there is such a mechanical procedure? – Kolmin Sep 11 '17 at 18:59
• If by mechanical procedure, you mean going from the inequation to the solution without having to make any test, I don't think so. But remember a test is a perfectly legitimate part of an algorithm. – Bernard Sep 11 '17 at 19:04
• Actually, what I mean is something that can be performed without thinking. That is, if we solve an equation of the form $ax^2 + bx = 0$, there is a sense in which we do not have to think. This applies to a lot of inequalities as well. But here it seems that you have to actually see what is going on. If I don't think (and I am rather good at that... ;D ), even if I proceed as in your edit, case (1) ends up "behaving" as follows: I have the two basic requirements $x \geq 1/2$ and $x <2$ and then I solve and get $x<1$ and $x>5$, which doesn't work. [Of course, correct me where I am wrong] – Kolmin Sep 11 '17 at 19:27

Let $$f(x)=\frac{\sqrt{2x-1}-(x-2)}{x-2},$$ $x\ge\frac12$ and $x\not=2$

$$f(x)=0 \Leftrightarrow \sqrt{2x-1}-(x-2)=0 \Leftrightarrow x=5$$ Then $f(x)<0$ if $\frac12 \le x<2$ or $x>5$

Answer: $\frac12 \le x<2$ or $x>5$

• I didn't understand what you did :( – danny Sep 6 '16 at 18:13
• @danny: I edited – Roman83 Sep 6 '16 at 18:26