# How to solve an irrational inequality?

How to solve the following inequality:

$$\sqrt{1-2x} < \sqrt{4 - x}$$

I don't understand why "$(1-2x)$ have to be $\ge 0$". If it was the rule for numbers inside a square root, I was checking whether we had to solve for "$4-x\ge 0$" as well.

• Square both sides of the inequality. Jul 19, 2014 at 17:09
• Please show what you have tried already. Jul 19, 2014 at 17:09
• I'm new. I'm really sorry. Thanks for reminding ^^ Jul 19, 2014 at 23:44
• $1-2x$ has to be $\geq 0$ because that is when the square root is defined. If it is $< 0$, you get a negative number inside a square root, which is undefined Jul 20, 2014 at 4:28
• So then, would it also work if I have 4−x≥0? That would give me a different answer though. Not 1/2 but x≤4 instead. Jul 20, 2014 at 16:03

\begin{align} \sqrt{1-2x}&<\sqrt{4-x} \\ \left(\sqrt{1-2x}\right)^2&<\left(\sqrt{4-x}\right)^2 \\ 1-2x&<4-x \\ -x&<3 \\ x&>-3 \\ \end{align} Remember the domain: $x\leq\frac 12$. Any $x$ greater than $\frac 12$ is not defined. Therefore the answer is: $$-3<x\leq \frac 12$$
You can solve this graphically as well. $$\sqrt{1-2x}<\sqrt{4-x}$$ $$\implies \sqrt{1-2x}-\sqrt{4-x}<0$$ I will now graph the function $f(x)=\sqrt{1-2x}-\sqrt{4-x}$ The solution to the inequality are the values less than $0$. We can see that these values are in the region $-3<x\leq \frac 12$.
First of all notice that for the L.H.S. to make sense, we should have $$x \le \frac{1}{2}$$ and for the R.H.S, we should have $$x \le {4}$$ now taking the square of both sides, we get $$1 - 2x < 4 - x$$ the above inequalities could be summarized as $$- 3 < x \le \frac{1}{2}$$