How do you solve #|x - 2| < | 2x + 1|#?

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Answer 1 · 2018-07-10T11:55:17

Solution: # x < 3 and x > 1/3 or (-oo, 3)uu(1/3,oo)#

#abs(x-2) < abs(2 x+1)# squaring both sides ,

#(x-2)^2 < (2 x+1)^2# or

#x^2-4 x+4 < 4 x^2 +4 x +1# or

#4 x^2 +4 x +1> x^2-4 x+4 # , transposing,

#4 x^2 +4 x +1 - x^2+4 x-4 >0 # or

#3 x^2 +8 x -3 >0 # or

#3 x^2 +9 x - x-3 >0 # or

#3 x( x+3)-1(x+3) >0 # or

#( x+3)(3 x-1) >0 #, critical points are # x=-3, x=1/3

# f(x)=( x+3)(3 x-1) #

Sign chart: When #x <-3 # sign of #f(x)# is #(-)*(-)=(+) :. >0#

When #-3 < x <1/3 # sign of #f(x)# is #(+)*(-)=(-) :. <0#

When #x > 1/3 # sign of #f(x)# is #(+)*(+)=(+) :. >0#

Solution:# x < 3 and x > 1/3 or (-oo, 3)uu(1/3,oo)# [Ans]