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DochEvi [55]
2 years ago
13

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

Mathematics
1 answer:
Elena-2011 [213]2 years ago
7 0

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\}

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