# How do you solve 32x-48<(32)/(x)?

Sep 19, 2015

solve $f \left(x\right) = 32 x - 48 < \frac{32}{x}$

Ans: $\left(- \frac{1}{2} , 2\right)$

#### Explanation:

$f \left(x\right) = 32 {x}^{2} - 48 x < 32$. Simplify by 16:
$f \left(x\right) = 2 {x}^{2} - 3 x - 2 < 0$ (1)
First, solve the quadratic equation $f \left(x\right) = 2 {x}^{2} - 3 x - 2 = 0.$
Transformed equation $f ' \left(x\right) = {x}^{2} - 3 x - 4 = 0$ (2). Factor pairs of (-4) --> (-1. 4). This sum is 4 - 1 = 3 = -b. Then the 2 real roots of (2) are -1 and 4. Back to the original equation (1). The 2 real roots are : $x = - \frac{1}{2}$ and $x = \frac{4}{2} = 2$.
Next, solve the inequality f(x) < 0. Between the 2 real roots $\left(- \frac{1}{2}\right)$ and (2), f(x) is negative (< 0), as opposite in sign to the + sign of a = 2.
Answer by open interval:$\left(- \frac{1}{2} , 2\right)$