Question

If it is correct i will mark as brainlist or i will report​

Answer 1

Answer:

The number of non-positive integral values of $k$ are contained in the following set:

$S_{k} = \{-1, 0\}$

Step-by-step explanation:

Let be the following second order polynomial:

$x^{2}-(k + 1)\cdot x + (k^{2}+k -8) = 0$, $k \le 0$ (1)

Whose roots can be found by the Quadratic Formula:

$x_{1,2} =\frac{k + 1\pm \sqrt{k^{2}+2\cdot k +1 - 4\cdot k^{2}-4\cdot k +32 }}{2}$

$x_{1,2} = \frac{k+1}{2} \pm \frac{\sqrt{33-2\cdot k -3\cdot k^{2}}}{2}$

Based on the statement, we have the following system of inequations:

$\frac{k+1}{2} + \frac{\sqrt{33 - 2\cdot k -3\cdot k^{2}}}{2} > 2$ (2)

$\frac{k+1}{2} - \frac{\sqrt{33 - 2\cdot k -3\cdot k^{2}}}{2} < 2$ (3)

By (2) we have:

$k + 1 + \sqrt{33-2\cdot k -3\cdot k^{2}} > 4$

$\sqrt{33 -2\cdot k - 3\cdot k^{2}} > 4 - (k + 1)$

$33 - 2\cdot k -3\cdot k^{2} > [4 - (k+ 1)]^{2}$

$33 - 2\cdot k -3\cdot k^{2} > 16 -8\cdot (k+1)+(k+1)^{2}$

$33 - 2\cdot k - 3\cdot k^{2} > 16-8\cdot k -8 + k^{2}+2\cdot k + 1$

$33 - 2\cdot k -3\cdot k^{2} > 9 -6\cdot k + k^{2}$

$0 > 4\cdot k^{2}-4\cdot k -24$

$4\cdot k^{2}-4\cdot k -24< 0$

$4\cdot (k^{2}-k-6) < 0$

$k^{2}-k - 6 < 0$

$(k -3)\cdot (k+2) < 0$

The solution is:

$k \in (-2, 3)$

Likewise, we get the following expression from (3):

$k^{2}-k - 6 > 0$

$(k -3)\cdot (k + 2) > 0$

The solution is:

$k \in (-\infty, -2)\,\cup\,(3, +\infty)$

The number of non-positive integral values of $k$ are contained in the following set:

$S_{k} = \{-1, 0\}$