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).

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step-by-step explanation:

b. hoped i'm right lol

step-by-step explanation: