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

Jul 9, 2016

(-inf., $\frac{3}{4}$] and [$\frac{4}{3}$, inf.)

#### Explanation:

$f \left(x\right) = 12 {x}^{2} - 25 x + 12 \ge 0$
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: $\frac{9}{12} = \frac{3}{4}$ and $\frac{16}{12} = \frac{4}{3}$.
Now, plot the 2 roots $\left(\frac{3}{4}\right)$ and $\left(\frac{4}{3}\right)$ on the number line. Since a > 0, the parabola opens upward. The function f(x) < 0 inside the interval $\left(\frac{3}{4} , \frac{4}{3}\right)$ and f(x) is positive outside this interval.
Solution set by intervals, $f \left(x\right) \ge 0$
Half closed intervals (-inf., $\frac{3}{4}$] and [$\frac{4}{3}$, inf.)
The 2 end points are included in the solution set.
Graph:

============= 0===== $\frac{3}{4}$------------ $\frac{4}{3}$ ===================