SOLUTION: given the equation x^2 +kx +2=0 find all possible values of k for one real solution, 2 real solutions, and no real solutions

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Question 348125: given the equation x^2 +kx +2=0 find all possible values of k for one real solution, 2 real solutions, and no real solutions
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
In the general form of the quadratic equaiton, ax%5E2+%2B+bx+%2B+c+=+0, the value of the discriminant, b%5E2+-+4ac, will indicate the number of real solutions:
  • b%5E2+-+4ac+=+0 means one real solution.
  • b%5E2+-+4ac+%3E+0 means two real solutions.
  • b%5E2+-+4ac+%3C+0 means no real solutions.

Your discriminant is:
k%5E2+-+4%281%29%282%29+=+k%5E2+-+8
So if k%5E2+-+8+=+0 you get one real solution. Solving this for k will tell us the k's that will result in one real solution. Adding 8 to each side we get:
k%5E2+=+8
This gives us
k+=+sqrt%288%29+=+2sqrt%282%29 or k+=+-2sqrt%282%29
For either of these values of k, there will be one real solution.

For two real solutions we want k%5E2+-+8+%3E+0. This means we need abs%28k%29+%3E+2sqrt%282%29. In other words, k+%3E+sqrt%282%29 or k+%3C+-sqrt%282%29.

For no real solutions we want k%5E2+-+8+%3C+0. This means we need abs%28k%29+%3C+2sqrt%282%29. In other words, k+%3C+sqrt%282%29 and k+%3E+-sqrt%282%29.