How do you solve #9x^2+3x-2>=0#?

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Answer 1 · 2017-01-26T19:17:32

#x=1/3>0#

#9x^2+3x-2-<0#

#:.=(3x-1)(3x+2)=0#

#:.3x-1=0#

#:.3x=1#

#:.x=1/3#

sustitute #x=1/3#

#:.9(1/3)^2+3(1/3)-2=0#

#:.9 xx 1/9+3(1/3)-2=0#

#:.9/9+3/3-2=0#

#:.1+1-2=0#

#3x+2=0#

#:.3x=-2#

#:.x=-2/3#

substitute #x=-2/3#

#:.9(-2/3)^2+3(-2/3)-2#

#:.9(-4/9)+3(-2/3)-2#

#:.-36/9-6/3-2=0#

#:.-4-2-2=0#

#:.-8=0# an extraneous solution