How do you solve #4( x ^ { 2} - 1) < 4x - 1#?

2 Answers
Καδήρ Κ.
Aug 28, 2017

#-1/2< x<3/2#

Explanation:

#4(x^2-1)<4x-1iff4x^2-4<4x-1iff#

#4x^2-4x-3<0iff4(x+1/2)(x-3/2)<0iff#

#(x+1/2)(x-3/2)<0iff-1/2< x<3/2#

Gazza
Aug 28, 2017

#-1/2 < x < 3/2#

Explanation:

#4 (x^2 - 1) < 4x - 1#
#4x^2 - 4 = 4x - 1#
#4x^2 - 4x - 4 + 3 = 0#
#4x^2 - 4x - 3 = 0#
Factorize
#(2x + 1)(2x - 3) = 0#
#2x + 1 = 0 or 2x - 3 = 0#
#x = (-1)/2 or x = 3/2#
We have
#x < (-1)/2# which doesn't work in original inequality
#(-1)/2 < x < 3/2# which works in original inequality
#x > 3/2# it doesn't work in original inequality

Thus,
The final answer is
#(-1)/2 < x < 3/2#

I hope I helped!

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