Answer:
The number of non-positive integral values of [tex]k[/tex] are contained in the following set:
[tex]S_{k} = \{-1, 0\}[/tex]
Step-by-step explanation:
Let be the following second order polynomial:
[tex]x^{2}-(k + 1)\cdot x + (k^{2}+k -8) = 0[/tex], [tex]k \le 0[/tex] (1)
Whose roots can be found by the Quadratic Formula:
[tex]x_{1,2} =\frac{k + 1\pm \sqrt{k^{2}+2\cdot k +1 - 4\cdot k^{2}-4\cdot k +32 }}{2}[/tex]
[tex]x_{1,2} = \frac{k+1}{2} \pm \frac{\sqrt{33-2\cdot k -3\cdot k^{2}}}{2}[/tex]
Based on the statement, we have the following system of inequations:
[tex]\frac{k+1}{2} + \frac{\sqrt{33 - 2\cdot k -3\cdot k^{2}}}{2} > 2[/tex] (2)
[tex]\frac{k+1}{2} - \frac{\sqrt{33 - 2\cdot k -3\cdot k^{2}}}{2} < 2[/tex] (3)
By (2) we have:
[tex]k + 1 + \sqrt{33-2\cdot k -3\cdot k^{2}} > 4[/tex]
[tex]\sqrt{33 -2\cdot k - 3\cdot k^{2}} > 4 - (k + 1)[/tex]
[tex]33 - 2\cdot k -3\cdot k^{2} > [4 - (k+ 1)]^{2}[/tex]
[tex]33 - 2\cdot k -3\cdot k^{2} > 16 -8\cdot (k+1)+(k+1)^{2}[/tex]
[tex]33 - 2\cdot k - 3\cdot k^{2} > 16-8\cdot k -8 + k^{2}+2\cdot k + 1[/tex]
[tex]33 - 2\cdot k -3\cdot k^{2} > 9 -6\cdot k + k^{2}[/tex]
[tex]0 > 4\cdot k^{2}-4\cdot k -24[/tex]
[tex]4\cdot k^{2}-4\cdot k -24< 0[/tex]
[tex]4\cdot (k^{2}-k-6) < 0[/tex]
[tex]k^{2}-k - 6 < 0[/tex]
[tex](k -3)\cdot (k+2) < 0[/tex]
The solution is:
[tex]k \in (-2, 3)[/tex]
Likewise, we get the following expression from (3):
[tex]k^{2}-k - 6 > 0[/tex]
[tex](k -3)\cdot (k + 2) > 0[/tex]
The solution is:
[tex]k \in (-\infty, -2)\,\cup\,(3, +\infty)[/tex]
The number of non-positive integral values of [tex]k[/tex] are contained in the following set:
[tex]S_{k} = \{-1, 0\}[/tex]
