Animated Solution for Mathematics - Basic Mathematics: The number of elements in the set $\{n\in \mathbb{Z}:|n^2-10n+19|<6\}$ is

The problem requires finding all integer values of $n$ that satisfy the inequality $|n^2-10n+19|<6$.

Geometrically, an absolute value inequality of the form $|x|<a$ implies that the expression $x$ is trapped between $-a$ and $a$. Therefore, our inequality is equivalent to the compound inequality:

This requires the parabola to simultaneously satisfy two conditions: it must be below the line $y=6$ and above the line $y=-6$.

Taming the Parabola

We first solve the upper boundary condition: $n^2-10n+19<6$. Subtracting $6$ from both sides yields:

To find the valid range for $n$, we determine the roots of the quadratic equation $n^2-10n+13=0$ using the quadratic formula:

Substituting the coefficients $a=1,b=-10,c=13$:

Given that $\sqrt{3}\approx 1.732$, we have $2\sqrt{3}\approx 3.464$. The roots are approximately $1.536$ and $8.464$. The integers $n$ satisfying this condition are $\{2,3,4,5,6,7,8\}$.

The Hidden Perfect Square

Next, we address the second condition: $n^2-10n+19>-6$. Adding $6$ to both sides results in:

Recognizing this as a perfect square, we rewrite the inequality as:

Since the square of any real number is non-negative, this inequality holds for all real numbers except where the expression equals zero. Thus, we must exclude the case where $n-5=0$, which implies $n eq 5$.

We combine our findings from the two conditions. The first condition provided the candidate set $\{2,3,4,5,6,7,8\}$, while the second condition imposed the constraint $n eq 5$.

Removing $5$ from the candidate set leaves us with $\{2,3,4,6,7,8\}$.

Counting these elements, we find there are exactly $6$ integer values of $n$ that satisfy the original inequality.