Theory of Intelligence • First-Principles AI & STEM Education

For any real solution to exist, the right-hand side must satisfy $x^2 - x + 1 \ge 0$ . Notice that:

x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} \ge \frac{3}{4} > 0 \quad (\forall x \in \mathbb{R})

Since the RHS is strictly positive for all real $x$ , no potential real solution will be rejected due to a negative RHS.

An equation $|A| = B$ splits into:

A = B \quad \text{or} \quad A = -B

\begin{aligned} 3x^2 - 7x + 2 &= x^2 - x + 1 \\ 2x^2 - 6x + 1 &= 0 \end{aligned}

Using the quadratic formula:

x = \frac{-(-6) \pm\sqrt{(-6)^2 - 4(2)(1)}}{2(2)} = \frac{6 \pm\sqrt{36 - 8}}{4} = \frac{6 \pm\sqrt{28}}{4} = \frac{6 \pm 2\sqrt{7}}{4} = \frac{3 \pm\sqrt{7}}{2}

Numerically: $x \approx \frac{3 - 2.646}{2} \approx 0.177$ and $x \approx \frac{3 + 2.646}{2} \approx 2.823$ .

\begin{aligned} 3x^2 - 7x + 2 &= -(x^2 - x + 1) \\ 3x^2 - 7x + 2 &= -x^2 + x - 1 \\ 4x^2 - 8x + 3 &= 0 \end{aligned}

Factor the quadratic trinomial:

\begin{aligned} (2x - 1)(2x - 3) &= 0 \\ x = \frac{1}{2} = 0.5 \quad &\text{or} \quad x = \frac{3}{2} = 1.5 \end{aligned}

All four solutions are real and yield valid equality:

x \in \left\{ \frac{1}{2}, \; \frac{3}{2}, \; \frac{3 - \sqrt{7}}{2}, \; \frac{3 + \sqrt{7}}{2} \right\}