Question

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

Answer 1

(-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` ===================