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

1 Answer
Sep 19, 2015

Answer:

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)#