Solve the inequality |a - 1| < |2a|.

Step-by-step solution

1. 1

   Justify squaring. Both sides are absolute values, hence both non-negative, and $t\mapsto t^2$ is strictly increasing on $[0,\infty)$. So squaring is an equivalence, not just an implication — no extraneous solutions can appear:

   $(a-1)^2<(2a)^2$

   This is the main advantage over case-splitting four sign combinations.

2. 2

   Expand and collect on one side.

   $a^2-2a+1<4a^2\;\Longrightarrow\;0<3a^2+2a-1$

   so we must solve $3a^2+2a-1>0$.

3. 3

   Factor the quadratic. Looking for factors of $3\cdot(-1)=-3$ summing to $2$ gives $3$ and $-1$:

   $3a^2+3a-a-1=3a(a+1)-1(a+1)=(3a-1)(a+1)$

   so the critical points are $a=-1$ and $a=\tfrac13$.

4. 4

   Use the upward parabola to read off the sign. The leading coefficient $3$ is positive, so $(3a-1)(a+1)$ is positive outside the roots and negative between them:

   $a<-1\qquad\text{or}\qquad a>\frac{1}{3}$

5. 5

   Test one point in each of the three regions. At $a=-2$: $|-3|=3<|-4|=4$ $\checkmark$. At $a=0$: $|-1|=1<|0|=0$ is false $\checkmark$ (correctly excluded). At $a=1$: $|0|=0<|2|=2$ $\checkmark$. The endpoints are excluded because there the two sides are equal.