Russ P. answered • 11/25/14

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Nina,

This is a quadratic equation whose solution is given by the quadratic formula, where A=3, B=k, and C=12.

The formula gives the solutions as:

x = {-B ± √(B

^{2}- 4AC)} /2AIt is the square root portion that determines how many and what kind of distinct solutions you will get.

So (B

^{2}- 4AC) = [k^{2}- 4(3)(12)] = [k^{2}- 144](a) For one real solution, both solutions must be identical, so the square root must be zero, and

[k

^{2}- 144] = 0. Hence k = ± 12. And x = -B/2A = -k/6. When k = +12, x

_{1}=x_{2 }= -2 and when k = -12 then x_{1}=x_{2}= +2.(b) For two imaginary solutions, you must have a negative non-zero real number inside the square root.

So [k

^{2}- 144] < 0 and -12 < k < 12. Then x = {-k ± i √(144 - k^{2})} /6 since bringing the imaginary i outside the square root makes the square root positive given the limits on k. (Recall that √-2 = √(-1)(+2) = i √+2).

(c) For two real solutions, you must have a positive non-zero real number inside the square root. This will happen when

|k| > 12. Then x = {-k ± √(k

^{2}- 144)} /6.BTW, remember that a quadratic polynomial has at most 2 distinct solutions.