Question

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

Answer 1

solve `f(x) = 32x - 48 < 32/x`

Ans: `(-1/2, 2)`

Explanation:

`f(x) = 32x^2 - 48x < 32`. Simplify by 16:
`f(x) = 2x^2 - 3x - 2 < 0` (1)
First, solve the quadratic equation `f(x) = 2x^2 - 3x - 2 = 0.`
Transformed equation `f'(x) = x^2 - 3x - 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 = -1/2` and `x = 4/2 = 2`.
Next, solve the inequality f(x) < 0. Between the 2 real roots `(-1/2)` and (2), f(x) is negative (< 0), as opposite in sign to the + sign of a = 2.
Answer by open interval:`(-1/2, 2)`