1. ## Inequality

Hi guys,

for some reason this equation is giving me hard time. There shouldn't be a solution to it....and I do get one.

My mistake is somewhere in the first line, because Maple makes the same mistake if I let it start from my second line of solution.

My first line is to power each side, so I do ()^2 for each side, meaning that on the left side the square root is gone, and on the right side, I can something like (a-b)^2.

Somehow it doesn't work, can someone solve it please ?

2. The inequality is equivalent to:

$\sqrt[]{1-2x}+2<\sqrt[]{2-2x}$

Taking squares and smplifying:

$4\;\sqrt[]{1-2x}<-3$

This is absurd because

$4\;\sqrt[]{1-2x}\;\geq 0$

Regards.

---
Fernando Revilla

3. thank you,
why did you move the "2" before you took square ?
isn't it allowed to take square when the statement is like (a-b)^2 and not (a+b)^2 ?
because if you do take square without moving it, you get an answer...

4. Originally Posted by WeeG
thank you,
why did you move the "2" before you took square ?
isn't it allowed to take square when the statement is like (a-b)^2 and not (a+b)^2 ?
because if you do take square without moving it, you get an answer...
I understand your question. First of all, when we write $\sqrt{a}$ we are considering the positive root. It is not necessary to move the "2", but the solution is more complicated.

The original inequality is:

$\sqrt{1-2x}<\sqrt{2-2x}-2 \qquad [1]$

Tahing into account that $\sqrt{1-2x}\geq 0$, necessary:

$\sqrt{2-2x}-2>0\qquad [2]$

If you solve $[2]$, you will obtain:

$x<-1\qquad [3]$

If you take squares in $[1]$ without moving the "2" you will obtalin:

$x\geq 7/32$

and this is a contradiction with $[3]$.

Regards.

---
Fernando Revilla