Suppose the real numbers a, b, and c are such that the graph of y = ax^2 + bx + c is entirely below the x-axis. What can you conclude about the solution(s) of the quadratic equation ax^2 + bx + c = 0?
A Only one solution, which is real
B Two real solutions, which are negative
C Two real solutions, one positive and one negative
D Two complex solutions which are not conjugates of each other
E There are no solutions
F Two complex solutions, which are conjugates of each other
2007-05-25 09:02:50 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
F. The solutions are where the graph crosses the X axis, so if it never does, all of the solutions are complex. In order to have real values for a, b, and c, the complex solutions must be conjugates of each other.
2007-05-25 09:09:31 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Since the graph never crosses the x-axis, there are no real solutions.
The solution of every quadratic equation is: x=(-b±(d)^(1/2))/2a and d=b^2-4*a*c
Only d will be complex
==>-b/2a is real and ±(d^0.5)/2a will cause the conjugates
so the solution will be two complex solutions, which are conjugates of each other.
2007-05-25 09:14:28 · answer #2 · answered by Johan 2 · 0⤊ 0⤋
F. You can also conclude that a is negative, and that b^2-4ac <0.
2007-05-25 09:13:07 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
B.
I think...
2007-05-25 09:05:18 · answer #4 · answered by Nathan 4 · 0⤊ 2⤋