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I am trying to solve $\sqrt{4m^2-4m+1}+|1-2m|\leq2$. Since i know $|1-2m| = \pm(1-2m)$ i tough solving $\sqrt{4m^2-4m+1}+1-2m\leq2$ and $\sqrt{4m^2-4m+1}-1+2m\leq2$. As solutions i get $0\leq2$ and $m\leq1$.

Is $0\leq2$ a solution or did i do something wrong? How should i proceed in solving this equation?

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Hint: $\sqrt{4m - 4m + 1} = \sqrt{(2m- 1)^2} = |2m - 1|$. – JavaMan Nov 23 '12 at 19:39
up vote 4 down vote accepted

First note that $$\sqrt{4m^2 - 4m + 1} = \sqrt{(1-2m)^2} = \left\vert 1-2m \right\vert$$ Hence, $$2 \geq \sqrt{4m^2 - 4m + 1} + \left\vert 1-2m \right\vert =2 \left\vert 1-2m \right\vert$$ This gives us that $$\left\vert 1-2m \right\vert \leq 1$$ Recall that $\vert x \vert \leq a$ is equivalent to $-a \leq x \leq a$.

Move your mouse over the gray area for the final answer.

Hence, we have that$$-1 \leq 1-2m \leq 1 \implies -2 \leq -2m \leq 0 \implies 0 \leq 2m \leq 2 \implies m \in [0,1]$$

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HINT: Note that $\sqrt{4m^2-4m+1}=\sqrt{(2m-1)^2}=|2m-1|=|1-2m|$, so you're really trying to solve $2|1-2m|\le 2$, i.e., $|1-2m|\le1$.

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