Absolute value and roots

Question

I've been trying to solve this problem and I always get 1, but the answer is $1 - 2x$.

If $x<\frac12$ then what is $\left|x-\sqrt{(x-1)^2}\right|$ ?

Answer 1

Note that $\sqrt {t^2}=|t|$ for all real numbers $t$. By the condition $x<\frac12$, it is clear that $x-1$ is negative, hence $\sqrt{(x-1)^2}=|x-1|=1-x$ and the final result is $|x-(1-x)|=|2x-1|=1-2x$ (because $2x-1<0$).

Answer 2

Note that if $c\ge 0$, the square root of $c$ is defined as the non-negative number whose square is $c$.

Thus $\sqrt{a^2}=|a|$. You can check this with for example $a=-3$. We have $\sqrt{(-3)^2}=\sqrt{9}=3$. So $\sqrt{(-3)^2}\ne -3$.

In particular, if $x\lt 1$, then $\sqrt{(x-1)^2}=|x-1|=1-x$.

Answer 3

The fundamental thing you have to use is $$ \sqrt{a^2}=|a| $$ So your expression is $$ \bigl|x-|x-1|\bigr|=\bigl|x-(1-x)\bigr|=\bigl|2x-1\bigr|=1-2x $$ Indeed $x<\frac{1}{2}$ entails $x<1$, so $|x-1|=1-x$. Also $2x-1<0$, so $|2x-1|=1-2x$.