Question

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

Answer 1

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]}