If y=x^2+kx-k for what values of k will the quadratic have 2 real solutions

The key to answering this question lies in finding the discriminant b^2-4ac. If b^2-4ac is greater than zero, the quadratic equation will have 2 real solutions.

Typing out y=x^2+kx-k, we see that a=1, b=k and c = - k. The discriminant in this case is

b^2 - 4 (a) (c), or k^2 - 4 (1) (-k), or k^2 + 4k

For which k is this k^2 + 4k = 0? Factor k^2 + 4k, obtaining k (k+4) = 0; the roots of this are k=0 and k = - 4.

Now set up a number line for k, drawing empty circles at k = 0 and k = - 4. These two values divide the number line into three intervals:

(-infinity, - 4), (-4, 0) and (0, infinity). Choose a test number for each interval:

-6 for the first, - 2 for the second and 4 for the third. Determine whether k^2 + 4k > 0 for any of these test numbers:

(-6) ^2 + 4 (-6) = 36 - 24 = + 12. Yes, the discriminant is + on the interval (-infinity, - 4)

(-2) ^2 + 4 (-2) = 4 - 8 = - 4. No, the discriminant is not + on (-4, 0).

(4) ^2 + 4 (4) = 32. Yes, the discrim. is + on (0, infinity).

So: The given quadratic will have two different, real roots on the two intervals (-infinity, - 4) and (0, infinity).