求方程 $|x^2 - 7x + 3| = 13 - x$ 的精确根。 在同一坐标系中，草绘方程为 $y = |x^2 - 7x + 3|$ 和 $y = 13 - x$ 的曲线，并标明所有交点的 $x$ 坐标值。由此精确求解不等式 $|x^2 - 7x + 3| < 13 - x$。

1. Formulate the two possible cases for the modulus equation.

$|x^2 - 7x + 3| = 13 - x$

Case 1:
$x^2 - 7x + 3 = 13 - x$
$x^2 - 6x - 10 = 0$

Case 2:
$x^2 - 7x + 3 = -(13 - x)$
$x^2 - 7x + 3 = -13 + x$
$x^2 - 8x + 16 = 0$

2. Solve the first quadratic equation using the quadratic formula to find the exact roots.

For $x^2 - 6x - 10 = 0$:
$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-10)}}{2(1)}$
$x = \frac{6 \pm \sqrt{36 + 40}}{2}$
$x = \frac{6 \pm \sqrt{76}}{2}$
$x = \frac{6 \pm 2\sqrt{19}}{2} = 3 \pm \sqrt{19}$

3. Solve the second quadratic equation by factorisation.

For $x^2 - 8x + 16 = 0$:
$(x - 4)^2 = 0$
$x = 4$

4. Verify the validity of the roots against the condition for the modulus.

Since a modulus expression must evaluate to a non-negative value, we require $13 - x \ge 0$, which implies $x \le 13$.

• For $x = 3 + \sqrt{19} \approx 7.36$, $x \le 13$ is satisfied.
• For $x = 3 - \sqrt{19} \approx -1.36$, $x \le 13$ is satisfied.
• For $x = 4$, $x \le 13$ is satisfied.

All obtained roots are valid solutions.

1. Retrieve the exact roots from part (a) to find the $x$-coordinates of the intersections.

Solving $|x^2 - 7x + 3| = 13 - x$ (from part a):
Case 1: $x^2 - 7x + 3 = 13 - x \implies x^2 - 6x - 10 = 0 \implies x = 3 \pm \sqrt{19}$
Case 2: $x^2 - 7x + 3 = -(13 - x) \implies x^2 - 8x + 16 = 0 \implies (x - 4)^2 = 0 \implies x = 4$
The roots are $x = 3 - \sqrt{19}$, $x = 4$, and $x = 3 + \sqrt{19}$.

2. Sketch the curve $y = |x^2 - 7x + 3|$ and the line $y = 13 - x$ on the same axes.

The graph of $y = |x^2 - 7x + 3|$ is a parabola opening upwards with the section below the $x$-axis reflected above it, forming two cusps on the $x$-axis.
The graph of $y = 13 - x$ is a straight line with a $y$-intercept of $13$.
The intersections are indicated on the $x$-axis at $x = 3 - \sqrt{19}$, $x = 4$, and $x = 3 + \sqrt{19}$. Because the intersection equation $x^2 - 8x + 16 = 0$ produced a repeated root at $x = 4$, the line is tangent to the top of the reflected section of the curve at this point.

3. Use the sketch to solve the strict inequality $|x^2 - 7x + 3| < 13 - x$.

The inequality requires the curve $y = |x^2 - 7x + 3|$ to be strictly below the line $y = 13 - x$.
From the sketch, the curve is below the line between the outer intersections $x = 3 - \sqrt{19}$ and $x = 3 + \sqrt{19}$.
However, at the tangent point $x = 4$, the curve touches the line (i.e., they are equal). Since the inequality is strict ($<$), this point must be excluded.

4. State the final solution.

$3 - \sqrt{19} < x < 4 \quad \text{or} \quad 4 < x < 3 + \sqrt{19}$