Key Concepts and Formulas

• Absolute Value Inequality: For any real number $x$ and positive real number $a$, $|x| < a$ is equivalent to $-a < x < a$.
• Quadratic Inequalities: To solve $ax^2 + bx + c < 0$ (or $> 0$), find the roots of $ax^2 + bx + c = 0$. If $a > 0$, the quadratic is negative between the roots and positive outside the roots.
• Quadratic Formula: The roots of $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Apply the Absolute Value Property We are given the inequality $|n^2 - 10n + 19| < 6$. Using the property $|x| < a \iff -a < x < a$, we can rewrite this as: $-6 < n^2 - 10n + 19 < 6$ This compound inequality can be split into two simultaneous inequalities:

1. $n^2 - 10n + 19 > -6$
2. $n^2 - 10n + 19 < 6$

Step 2: Solve the First Inequality Consider the inequality $n^2 - 10n + 19 > -6$. Add 6 to both sides to set the inequality to zero: $n^2 - 10n + 19 + 6 > 0$ $n^2 - 10n + 25 > 0$ The left side is a perfect square trinomial: $(n-5)^2$. So, the inequality becomes: $(n-5)^2 > 0$ A squared real number is always non-negative. For $(n-5)^2$ to be strictly greater than 0, $(n-5)$ must not be zero. Therefore, $n-5 \neq 0$, which means $n \neq 5$. The integers satisfying this are all integers except 5.

Step 3: Solve the Second Inequality Consider the inequality $n^2 - 10n + 19 < 6$. Subtract 6 from both sides to set the inequality to zero: $n^2 - 10n + 19 - 6 < 0$ $n^2 - 10n + 13 < 0$ To find the values of $n$ for which this quadratic is negative, we first find the roots of the equation $n^2 - 10n + 13 = 0$ using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=1$, $b=-10$, and $c=13$. $n = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(13)}}{2(1)}$ $n = \frac{10 \pm \sqrt{100 - 52}}{2}$ $n = \frac{10 \pm \sqrt{48}}{2}$ Simplify $\sqrt{48}$: $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$. $n = \frac{10 \pm 4\sqrt{3}}{2}$ $n = 5 \pm 2\sqrt{3}$ The roots are $n_1 = 5 - 2\sqrt{3}$ and $n_2 = 5 + 2\sqrt{3}$. Since the coefficient of $n^2$ is positive ($a=1>0$), the parabola $y = n^2 - 10n + 13$ opens upwards. Thus, the quadratic is negative between its roots. The solution to $n^2 - 10n + 13 < 0$ is $5 - 2\sqrt{3} < n < 5 + 2\sqrt{3}$.

Step 4: Determine Integer Solutions from the Second Inequality We need to approximate the values of the roots to find the integers $n$ in the interval $(5 - 2\sqrt{3}, 5 + 2\sqrt{3})$. We know that $\sqrt{3} \approx 1.732$. So, $2\sqrt{3} \approx 2 \times 1.732 = 3.464$. The interval becomes: $5 - 3.464 < n < 5 + 3.464$ $1.536 < n < 8.464$ The integers $n$ that satisfy this inequality are $2, 3, 4, 5, 6, 7, 8$.

Step 5: Combine the Conditions We need to find the integers $n$ that satisfy both conditions:

• From Step 2: $n \in \mathbb{Z} \setminus \{5\}$ (all integers except 5).
• From Step 4: $n \in \{2, 3, 4, 5, 6, 7, 8\}$.

The intersection of these two sets is $\{2, 3, 4, 5, 6, 7, 8\} \setminus \{5\}$. This gives the set of integers $\{2, 3, 4, 6, 7, 8\}$.

Step 6: Count the Number of Elements The set of integers satisfying the original inequality is $\{2, 3, 4, 6, 7, 8\}$. The number of elements in this set is 6.

Common Mistakes & Tips

• Approximation Accuracy: Ensure sufficient decimal places are used when approximating square roots to correctly identify the integers within an interval. For $\sqrt{3}$, $1.732$ is generally adequate.
• Strict vs. Non-Strict Inequalities: Carefully distinguish between $<$ and $\le$ (or $>$ and $\ge$). Strict inequalities exclude boundary points, while non-strict inequalities include them.
• Intersection of Solutions: When solving a compound inequality like $-a < x < a$, the solution set is the intersection of the solution sets of the two individual inequalities.

Summary The problem required solving an absolute value inequality, which was transformed into two quadratic inequalities. The first inequality, $(n-5)^2 > 0$, was satisfied by all integers except $n=5$. The second inequality, $n^2 - 10n + 13 < 0$, was solved by finding its roots $5 \pm 2\sqrt{3}$ and determining the integers between them, which were $\{2, 3, 4, 5, 6, 7, 8\}$. Combining these conditions, we excluded $n=5$ from this list, resulting in the set $\{2, 3, 4, 6, 7, 8\}$, which contains 6 elements.

The final answer is \boxed{6}.