For a > 0, solve the inequality ax^2 - (2a - 1)x + 2 >= 2x.

Step-by-step solution

1. 1

   Collect everything on the left. Subtracting $2x$ merges the two linear terms:

   $ax^{2}-\left[(2a-1)+2\right]x+2\ge 0\ \Longrightarrow\ ax^{2}-(2a+1)x+2\ge 0$

2. 2

   Factor the quadratic. Try $(x-2)(ax-1)$ and expand to confirm:

   $(x-2)(ax-1)=ax^{2}-x-2ax+2=ax^{2}-(2a+1)x+2\ \checkmark$

   so the inequality is $(x-2)(ax-1)\ge 0$.

3. 3

   Normalise the leading coefficient. Because $a>0$, dividing by $a$ does not flip the inequality, and $ax-1=a\left(x-\tfrac1a\right)$:

   $(x-2)\left(x-\frac{1}{a}\right)\ge 0$

   The roots are $2$ and $\tfrac1a$; the parabola opens upward, so the solution is outside the roots — but which root is on the left depends on $a$.

4. 4

   Case $0<a<\tfrac12$ (so $\tfrac1a>2$). The roots in order are $2<\tfrac1a$, giving

   $\left(-\infty,2\right]\cup\left[\frac{1}{a},+\infty\right)$

   Check with $a=\tfrac14$ (roots $2$ and $4$): at $x=0$ the expression is $2\ge 0$ ✓, at $x=3$ it is $-\tfrac14<0$ ✗, at $x=5$ it is $\tfrac34\ge 0$ ✓.

5. 5

   Case $a=\tfrac12$ (so $\tfrac1a=2$). The roots coincide and the inequality becomes $\tfrac12(x-2)^{2}\ge 0$, true for every real number:

   $x\in\mathbb{R}$

6. 6

   Case $a>\tfrac12$ (so $\tfrac1a<2$). Now $\tfrac1a$ is the smaller root:

   $\left(-\infty,\frac{1}{a}\right]\cup\left[2,+\infty\right)$

7. 7

   Collect the three branches. The whole answer hinges on the single comparison $\tfrac1a$ versus $2$, i.e. on whether $a<\tfrac12$, $a=\tfrac12$ or $a>\tfrac12$ — the standard move for any parametric quadratic whose roots depend on the parameter.