Question

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

Show that 3k^3-3k-1=0​

Answer 1

(See explanation for further details)

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`