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

1 Answer
sente
Jul 23, 2016

#2x^2+3x>0 <=> x in (-oo, -3/2)uu(0,oo)#

Explanation:

The trick is to note that the sign of a polynomial expression can only be different at two points if the expression evaluates to #0# at some point between them. Then, if we consider the intervals on either side of where the expression evaluates to #0#, we can find all points where it is positive

Solving for where it is #0#:

#2x^2+3x=x(2x+3)=0#

# <=> x=0 or 2x+3=0#

# <=> x=0 or x= -3/2#

Thus the intervals in question are #(-oo, -3/2), (-3/2, 0), (0, oo)#

From here, we can just test a point in each to see what the sign of the expression will be:

#2(-2)^2+3(-2)=2(4)-6=8-6=2>0#

Thus #2x^2+3x>0# on #(-oo,-3/2)#

#2(-1)^2+3(-1)=2-3=-1 <0#

Thus #2x^2+3x<0# on #(-3/2,0)#

#2(1)^2+3(1) = 2+3 = 5>0#

Thus #2x^2+3x >0# on #(0,oo)#

Putting it together, then,

#2x^2+3x>0 <=> x in (-oo, -3/2)uu(0,oo)#

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