How do you solve #2x^2+3x>0# using a sign chart?

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Answer 1 · 2017-02-10T17:10:53

The answer is #x in ]-oo, -3/2[ uu ]0, +oo[#

Let #f(x)=2x^2+3x=x(2x+3)#

We now, build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-3/2##color(white)(aaaa)##0##color(white)(aaaa)##+oo#

#color(white)(aaaa)##2x+3##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x##color(white)(aaaaaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)>0# when #x in ]-oo, -3/2[ uu ]0, +oo[#