How do you find the zeros of #y=12x^2+8x-15#?

1 Answer
Apr 2, 2017

Answer:

Two real roots: #5/6 and - 3/2#

Explanation:

Solve y = 12x^2 + 8x - 15 = 0 by the new Transforming Method (Socratic Search)
Transformed equation y' = x^2 + 8x - 180 = 0
Compose factor pairs of (-180) --> ... (10, -18). This sum is -8 = -b.
The 2 real roots of y' are: 10 and - 18.
The 2 real roots of y are:
#x1 = 10/a = 10/12 = 5/6# and
#x2 = - 18/a = -18/12 = -3/2#