How do you solve # 12x^2 - 25x + 12 > =0#?

1 Answer
Jul 9, 2016

Answer:

(-inf., #3/4#] and [#4/3#, inf.)

Explanation:

#f(x) = 12x^2 - 25x + 12 >= 0#
First solve f(x) = 0 to find its 2 real roots, using the new Transforming Method (Google, Yahoo Search).
Transformed equation f'(x) = x^2 - 25x + 144 = 0.
Find 2 real roots, that have the same sign, knowing their sum (-b = 25) and their product (ac = 144).
Factor pairs of (ac = 144) --> (6, 24)(8, 18)(9, 16). This last sum is (25 = -b). Consequently, the 2 real roots of f'(x) are: 9 and 16.
Back to f(x), its 2 real roots are: #9/12 = 3/4# and #16/12 = 4/3#.
Now, plot the 2 roots #(3/4)# and #(4/3)# on the number line. Since a > 0, the parabola opens upward. The function f(x) < 0 inside the interval #(3/4, 4/3)# and f(x) is positive outside this interval.
Solution set by intervals, #f(x) >= 0#
Half closed intervals (-inf., #3/4#] and [#4/3#, inf.)
The 2 end points are included in the solution set.
Graph:

============= 0===== #3/4#------------ #4/3# ===================