How do you graph and solve #|4x + 3| ≥ 19#?

1 Answer
Dec 27, 2017

Answer:

#x in (-oo,-11/2]uuu[4,oo)#

Explanation:

First Method: Rewrite as quadratic function

#(4x+3)^2≥19^2#

#16x^2+24x-352≥0#

#2x^2+3x-44≥0#

#x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)#

#=>x_(1,2)=4; -11/2#

#x in (-oo,-11/2]uuu[4,oo)#

Second Method:

#|4x+3|≥19#

#4x_1+3=19#
#4x_1=16#
#x_1=4#

#-(4x_2+3)=19#
#-4x_2-3=19#
#-4x_2=22#
#x_2=22/-4=-11/2#

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(I personally prefer first method eventhough it is more difficult. I can tell the interval in a second)