How do you solve #3x^2 + 19x -14 = 0#?

1 Answer
Aug 11, 2015

Answer:

Solve #y = 3x^2 + 19x - 14 = 0 #(1)

Ans: #2/3# and -7.

Explanation:

I use the new Transforming Method (Google, Yahoo)
Transformed equation: #y' = x^2 + 19x - 52 = 0# (2) Roots have opposite signs (Rule of Signs).
Factor pairs of (-42) --> (-2, 21). This sum is 19 = b. Then the 2 real roots of (2) are the opposite. They are: y1 = 2 and y2 = - 21.
Back to original equation (1). The 2 real roots are: #x1 = (y1)/a = 2/3#, and #x2 = (y2)/a = -21/3 = -7.#