How do you solve #(x-2)(x+5)(2x^2+5x-3)>=0#?

1 Answer
Tamir E.
Nov 14, 2017

#x<=5#
or
#-3<=x<=0.5#
or
#x>=2#

Explanation:

#g_((x))=2x^2+5x-3#

#x_(1,2)=(-5+-sqrt(25-4*2*(-3)))/(2*2)#
#=(-5+-sqrt(25+24))/(4)#
#=(-5+-sqrt(49))/(4)#
#=(-5+-7)/(4)#
#x_1=(-5+7)/(4)=(2)/4=0.5#
#x_2=(-5-7)/(4)=(-6)/4=-3#

#g_((x))=2(x+3)(x-0.5)=(x+3)(2x-1)#

#(x-2)(x+5)(x+3)(2x-1)>=0#

#y=0# when:
#x=-5#
#x=-3#
#x=0.5#
#x=2#

we check for #x=0#:
#y_((0))=(-2)(5)(3)(-1)=(positive)#

therefore:
#(x<-5)#
#(-5<x<-3)#
#(-3<x<0.5) >0#
#(0.5<x<2)#
#(x>2)#

and so:
#(x<-5) >0#
#(-5<x<-3) <0#
#(-3<x<0.5) >0#
#(0.5<x<2) <0#
#(x>2) >0#

so the result is:

#x<=5 or -3<=x<=0.5 or (x>=2) #

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