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zysi [14]
3 years ago
6

Given that the quadratic equation has equal roots (k^2-1)x^2-2kx-3k-1=0

Mathematics
1 answer:
Alik [6]3 years ago
7 0

Answer:

(See explanation for further details)

Step-by-step explanation:

The real expression is:

(k^{2}-1)\cdot x{{2} - 2\cdot k \cdot x - 3\cdot k + 1 = 0

The general equation for the second-order polynomial is:

x = \frac{2\cdot k \pm \sqrt{4\cdot k^{2}-4\cdot (k^{2}-1)\cdot (-3\cdot k + 1)}}{k^{2}-1}

This condition must be observed for the case of a quadratic equation with equal roots:

4\cdot k^{2} - 4\cdot (k^{2}-1)\cdot (-3\cdot k + 1) = 0

k^{2} + (k^{2}-1)\cdot (3\cdot k + 1) = 0

k^{2} + 3\cdot k^{3} - 3\cdot k - k^{2}-1 = 0

3\cdot k^{3} - 3\cdot k - 1 = 0

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