Asked by Tish
How many solutions are there to the equation -0.3x^2+25x+16=0, how would you come to this conclusion?
Answers
Answered by
Henry
IF:
B^2 < 4AC: 2 Imaginary solutions.
B^2 = 4AC: 1 Real solution.
B^2 > 4AC: 2 Real solutions.
B^2 = (25)^2 = 625.
4AC = 4*0.30*16 = 19.2.
B^2 > 4AC, Therefore, we have 2 real solutions:
Use Quadratic Formula and get:
X = -0.64449.
X = -82.68834.
B^2 < 4AC: 2 Imaginary solutions.
B^2 = 4AC: 1 Real solution.
B^2 > 4AC: 2 Real solutions.
B^2 = (25)^2 = 625.
4AC = 4*0.30*16 = 19.2.
B^2 > 4AC, Therefore, we have 2 real solutions:
Use Quadratic Formula and get:
X = -0.64449.
X = -82.68834.
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