How do you solve #|5x+2|>=|3x-4|#?

1 Answer
Jun 12, 2017

Answer:

Solution : # x<= -3 or x>=1/4# OR # (-oo,-3]uu[1/4,oo)#

Explanation:

#|5x+2| >= |3x-4|# squaring both sides we get
# (5x+2)^2 >= (3x-4)^2 # or
#25x^2+20x+4 >= 9x^2 -24x +16# or
#25x^2+20x+4 - 9x^2 +24x -16 >=0# or
#16x^2+44x-12>=0# or
#4x^2+11x-3>=0# or
#(4x-1)(x+3).>=0#. Critical points are #x=-3 and x=1/4#
Sign Change:
For #x=-3 or x=1/4; (4x-1)(x+3)=0#
For #x<-3 ; (4x-1)(x+3)= (-*-)= (+) ; (4x-1)(x+3)>0 #

For #-3 < x < 1/4 ; (4x-1)(x+3)= (-*+)= (-) ; (4x-1)(x+3)<0 #

For #x>1/4 ; (4x-1)(x+3)= (+*+)= (+) ; (4x-1)(x+3)>0 #
So solution : # x<= -3 or x>=1/4# or # (-oo,-3]uu[1/4,oo)#
The graph also confirms the findings.
graph{4x^2+11x-3 [-40, 40, -20, 20]}