How do you find the zeros, real and imaginary, of y=9x^2-4x-15 using the quadratic formula?

1 Answer
Nghi N.
Jul 30, 2016

1.53 and - 1.09

Explanation:

y = 9x^2 - 4x - 15 = 0
Use the new quadratic formula in graphic form (Socratic Search).
D = d^2 = b^2 - 4ac = 16 + 540 = 556 --> d = +- 23.58
There are 2 real roots:
x = -b/(2a) +- d/(2a) = 4/18 +- 23.58/18 = (4 +- 23.58)/18
x1 = 27.58/18 = 1.53
x2 = - 19.58/18 = - 1. 09
graph{9x^2 - 4x - 15 [-40, 40, -20, 20]}

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