- #1
mathdad
- 1,283
- 1
Find values of k such that the equation has exactly one real root.
1. 3x^2 + (sqrt{2k})x + 6 = 0
2. kx^2 + kx + 1 = 0
Question:
Do the questions above involve the discriminant?
If so, I must apply b^2 - 4ac = 0, right?
1. 3x^2 + (sqrt{2k})x + 6 = 0
2. kx^2 + kx + 1 = 0
Question:
Do the questions above involve the discriminant?
If so, I must apply b^2 - 4ac = 0, right?