Complete the square for 3x^2 - 4x + 2, then find its zeros.

Step-by-step solution

1. 1

   Test for real factorisation before working. With $a=3$, $b=-4$, $c=2$:

   $\Delta=(-4)^{2}-4(3)(2)=16-24=-8$

   Because $\Delta<0$ the quadratic has no real roots, so it cannot be factored into real linear factors. Completing the square is the productive alternative — it yields the vertex and the complex roots in one go.

2. 2

   Factor the leading coefficient out of the $x$ terms only. Leaving the constant $2$ outside keeps the fractions smaller:

   $3x^{2}-4x+2=3\left(x^{2}-\tfrac43x\right)+2$

   Note $-4/3$, not $-4$: dividing the middle coefficient by $a=3$ is where sign and fraction slips usually happen.

3. 3

   Complete the square inside the bracket. Half of $-\tfrac43$ is $-\tfrac23$, and $\left(-\tfrac23\right)^{2}=\tfrac49$:

   $x^{2}-\tfrac43x=\left(x-\tfrac23\right)^{2}-\tfrac49$

   Substituting back and distributing the $3$ — remembering it multiplies the correction term too:

   $3\left(x-\tfrac23\right)^{2}-3\cdot\tfrac49+2=3\left(x-\tfrac23\right)^{2}-\tfrac43+2=3\left(x-\tfrac23\right)^{2}+\tfrac23$

4. 4

   Read off the vertex and minimum. Since $3\left(x-\tfrac23\right)^{2}\ge0$ with equality only at $x=\tfrac23$, the minimum value is $\tfrac23$ at $x=\tfrac23$. The identity was confirmed at $100$ exact rational points ✓, and the minimum is strictly positive — consistent with $\Delta<0$.

5. 5

   Find the zeros over the complex numbers. From the completed square, $3\left(x-\tfrac23\right)^{2}=-\tfrac23$, so $\left(x-\tfrac23\right)^{2}=-\tfrac29$ and

   $x=\frac{2}{3}\pm\frac{i\sqrt{2}}{3}=\frac{2\pm i\sqrt{2}}{3}$

   The quadratic formula agrees: $x=\frac{4\pm\sqrt{-8}}{6}=\frac{4\pm2i\sqrt{2}}{6}=\frac{2\pm i\sqrt{2}}{3}$. Substituting either root back gives a residual below $10^{-12}$ ✓.