Answer:
There are 15 different possible values of n:
{-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7}
Step-by-step explanation:
The given inequality is
[tex] {x}^{2} + nx + 15 \: < \: 0[/tex]
For to have no solution, the discriminant must be less than zero.
This means that:
[tex] {n}^{2} - 4 \times 1 \times 15 \: < \: 0 [/tex]
[tex]{n}^{2} - 60 \: < \: 0 [/tex]
[tex]{(n} - 4 \sqrt{15} )(n + 4 \sqrt{15} ) \: < \: 0 [/tex]
This implies that:
[tex] - 7.75 \: < \: n \: < \: 7.75[/tex]
The integer values are , {-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7}
We consider the graph of
\[y = x^2 + nx + 15,\]where $n$ is some constant. The graph is an upward-opening parabola, and the solutions of $x^2 + nx +15<0$ correspond to values of $x$ for which this parabola is below the $x$-axis. The parabola only goes below the $x$-axis if the quadratic $x^2 + nx + 15$ has two distinct roots. Otherwise, the parabola does not go below the $x$-axis. So, the inequality $x^2 + nx + 15<0$ has no solutions if and only if the quadratic $x^2+nx + 15$ does not have two distinct roots.
The quadratic $x^2 + nx + 15$ fails to have two distinct roots only when its discriminant, which is $n^2 - 4(1)(15)$, is nonpositive. So, we must have $n^2 - 60 \le 0$, or $n^2 \le 60$. This inequality holds for $-\sqrt{60} \le n \le\sqrt{60}$. Therefore, the integers $n$ for which $x^2 + nx + 15 < 0$ has no real solutions $x$ are the integers from $-7$ up to 7. There are $\boxed{15}$ such integers.
