Theory of Intelligence • First-Principles AI & STEM Education

Step-by-Step Solution

Step 1: Establish the Domain Restriction

Since the absolute value on the left-hand side is non-negative, a real solution requires the right-hand side to be non-negative:

x + 2 \ge 0 \implies x \ge -2

Step 2: Set Up the Double Inequality

For any non-negative RHS, $|A| \le B \iff -B \le A \le B$ :

-(x + 2) \le x^2 - 5x + 6 \le x + 2

This splits into a system of two simultaneous inequalities.

Step 3: Solve Inequality 1 (Upper Bound)

\begin{aligned} x^2 - 5x + 6 &\le x + 2 \\ x^2 - 6x + 4 &\le 0 \end{aligned}

Find the roots of the equation $x^2 - 6x + 4 = 0$ :

x = \frac{-(-6) \pm\sqrt{(-6)^2 - 4(1)(4)}}{2(1)} = \frac{6 \pm\sqrt{36 - 16}}{2} = \frac{6 \pm\sqrt{20}}{2} = 3 \pm\sqrt{5}

Numerically: $3 - \sqrt{5} \approx 0.764$ and $3 + \sqrt{5} \approx 5.236$ . Because the parabola opens upwards, it is non-positive between the roots:

x \in [3 - \sqrt{5}, \; 3 + \sqrt{5}]

Step 4: Solve Inequality 2 (Lower Bound)

\begin{aligned} x^2 - 5x + 6 &\ge -(x + 2) \\ x^2 - 5x + 6 &\ge -x - 2 \\ x^2 - 4x + 8 &\ge 0 \end{aligned}

Examine the discriminant of $x^2 - 4x + 8$ :

\Delta = (-4)^2 - 4(1)(8) = 16 - 32 = -16 < 0

Since $\Delta < 0$ and $a = 1 > 0$ , the quadratic $x^2 - 4x + 8$ is strictly positive for all $x \in \mathbb{R}$ . This condition is satisfied unconditionally for every real number!

Step 5: Intersect All Solution Sets

Intersecting:

$x \in [3 - \sqrt{5}, 3 + \sqrt{5}]$
$x \in \mathbb{R}$
$x \ge -2$ (note $3 - \sqrt{5} \approx 0.764 > -2$ , so this is fully respected)

The final solution interval is:

x \in [3 - \sqrt{5}, \; 3 + \sqrt{5}] \approx [0.764, \; 5.236]