Solve the inequality √(3x − 1) > 2x.

Step-by-step solution

1. 1

   Find the domain first. The radicand must be non-negative:

   $3x-1\ge 0\quad\Longrightarrow\quad x\ge\frac13$

   Everything that follows is restricted to $x\ge\tfrac13$.

2. 2

   Check whether the right-hand side can be negative. Squaring is only reversible when both sides are non-negative. On the domain $x\ge\tfrac13>0$ we have $2x>0$, so both sides are positive and squaring is a valid equivalence — no case split is needed.

3. 3

   Square both sides.

   $3x-1>4x^{2}\quad\Longrightarrow\quad 0>4x^{2}-3x+1$

4. 4

   Test the quadratic for negativity. Its discriminant is

   $(-3)^{2}-4(4)(1)=9-16=-7<0$

   and its leading coefficient $4$ is positive, so $4x^{2}-3x+1>0$ for every real $x$. The required inequality $4x^{2}-3x+1<0$ therefore has no solutions at all.

5. 5

   Conclude.

   $\boxed{\text{No solution: the solution set is empty}}$

6. 6

   Verify at a few points and see why. At $x=\tfrac13$: $\sqrt0=0$ versus $\tfrac23$ — the left side loses. At $x=\tfrac12$: $\sqrt{0.5}=0.707$ versus $1$ — loses. At $x=1$: $\sqrt2=1.414$ versus $2$ — loses. The line $2x$ grows faster than $\sqrt{3x-1}$ everywhere on the domain, and by the discriminant computation it is strictly above it even at the closest approach (near $x=\tfrac{25}{48}$, where the gap is smallest but still positive).