Question

solve for \(x\).

Answer 1

\(\large \color{black}{\begin{align} \sqrt{x^2}<x+1\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

Need help @mathmath333 ^.^

Answer 3

If you think im new and dont know much about OS this is my alt account .-. my original is @just_one_last_goodbye

Answer 4

Let's take a look... ^_^

Answer 5

but @mathmath333 I'm being monitored by mods and cant give a direct answer so please stay I need your participation

Answer 6

First we need to find the "real region" of the problem

Answer 7

\[\sqrt{x ^{2}}\] is real to what?

Answer 8

@mathmath333 u there?

Answer 9

what have you got math?

Answer 10

i have got \(x>-1/2\) and \(x>0\)

Answer 11

It is real to \[(x ^{2}) \ge 0\]

Answer 12

@mathmath333 thats the answer x > -1/2 is the solution

Answer 13

x>-1/2 covers x>0 also, so answer is x>-1/2

Answer 14

\[[Solution: x > -\frac{ 1 }{ 2 }\]
\[[Decimal: x > -0.5]\]
\[[Interval Notation: (-\frac{ 1 }{ 2 }, \infty) ]\]

Answer 15

Oh hey @wio huge fan of your extension that you made for OS ^.^ im currently using all of them

Answer 16

but the common region between \(x>0.5\) and \(x>0\) so why not \(x>0\)

Answer 17

x>-0.5 not x>0.5

Answer 18

yea i mean \(x>-0.5\)

Answer 19

You can split it up: \[
x\lt x+1 \\
-x\lt x+1
\]The first equation will be true for all real \(x\), and so only the second equation matters.
That equation leads to: \[
0\lt 2x+1 \implies -\frac 12 \lt x
\]