How do you solve # 3x^2 - 14x - 5 = 0#?

2 Answers
Aug 3, 2015

Answer:

#x_(1,2) = (14 +- 16)/6#

Explanation:

You could solve this quadratic equation by using the quadratic formula.

For a quadratic equation that takes the general form

#color(blue)(ax^2 + bx + c = 0)#

the two solutions can be determined by using the quadratic formula

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)#

In your case, #a=3#, #b=-14#, and #c=-5#. This means that the two solutions will be

#x_(1,2) = (-(-14) +- sqrt( (-14)^2 - 4 * 3 * (-5)))/(2 * 3)#

#x_(1,2) = (14 +- sqrt(256))/6#

#x_(1,2) = (14 +- 16)/6 = {(x_1 = (14 + 16)/6 = color(green)(5)), (x_2 = (14 - 16)/6 = color(green)(-1/3)) :}#

Aug 4, 2015

Answer:

Solve y = 3x^2 - 14x - 5 = 0 (1)

Ans: (- 1/3) and (5).

Explanation:

I use the new Transforming Method.
Transformed y' = x^2 - 14x - 15 = 0. (2)
Since (a -b + c = 0) --> 2 real roots of (2) are (-1) and (-c/a = 15).
Back to equation (1), the 2 real roots are: #(- 1/3)# and #(15/3 = 5)#