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?

2. ## Re: Quadratic equation help!!!

Originally Posted by yeatch
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?

3. ## Re: Quadratic equation help!!!

The integer replies that it should be a whole number not a fraction. I do not see any solutions to this problem...

4. ## Re: Quadratic equation help!!!

do i just chose any value for k as long as its uneven and odd ?

5. ## Re: Quadratic equation help!!!

Originally Posted by yeatch
do i just chose any value for k as long as its uneven and odd ?
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.

6. ## Re: Quadratic equation help!!!

Hey Jeff, ye olde mudder you; how you doin? Miss Lookagain?

7. ## Re: Quadratic equation help!!!

Originally Posted by Wilmer
Hey Jeff, ye olde mudder you; how you doin? Miss Lookagain?
Mon ami, je ne regrette point l'absence de mademoiselle lookagain. Nor mmmbot.

8. ## Re: Quadratic equation help!!!

Originally Posted by JeffM
Mon ami, je ne regrette point l'absence de mademoiselle lookagain. Nor mmmbot.