If kx^2-x-k-1=0 calculate the values of k where the roots of the quadratic are:
a.Real and uneven
b.Real and have the opposite sign..
So far i got x=-1 or x =1+1/k for the roots...how do i assess a and b?
$x = \dfrac{-\ (-\ 1) \pm \sqrt{(-1)^2 - 4k(-k - 1}}{2k} = \dfrac{1 \pm \sqrt{4k^2 + 4k + 1}}{2k} = \dfrac{1 \pm \sqrt{(2k + 1)^2}}{2k} \implies$
$x = \dfrac{-2k}{2k} = - 1\ or\ x = \dfrac{1 + 2k + 1}{2k} = \dfrac{2k + 2}{2k} = 1 + \dfrac{1}{k}.$
Well done so far.
Notice that the question asks about values (plural) of k. "Uneven" is a bit weird. Normally, one would say odd, which would imply an integer. If "uneven" means "not an even integer," what values of k are acceptable. If "uneven" means an odd integer, what values of k are acceptable. As for part b, notice that one root is negative, what does that imply about k if the other root is positive?
The condition imposed by the question involves the roots of the equation, not k itself. It is provable that no odd integer can be the root of that equation.
$Assume\ 1 + \dfrac{1}{k} = 2j + 1,\ where\ j \in \mathbb Z.$
$0 = x^2 - x - k - 1 = (x + 1)(x - 2j - 1) = x^2 + (1 - 2j - 1)x - 2j - 1 = x^2 - 2jx - 2j - 1 \implies - 2j = - 1 \implies j \not \in \mathbb Z.$
So, assuming you have transcribed the question correctly, there is NO VALUE of k that will make both roots of that equation odd integers.
But that still leaves the issue of whether there ARE values of k that will make one root positive and one root negative. Because one root is minus 1, this question boils down to what values of k will make the other root positive.
If you are still lost, please ask more questions.