How do you solve #6x^2+x>1#?

1 Answer
Sep 2, 2016

Answer:

#color(green)(x < -1/2)# or #color(green)(x > 1/3)#

Explanation:

#6x^2+x > 1# is equivalent to #6x^2+x-1 > 0#

Factoring #6x^2+x-1#
we have #(3x-1)(2x+1)#
which will be equal to zero when #x=1/3# or #x=-1/2#

We not interested in when #6x^2+x-1# is equal to zero.
Rather we are interested in the ranges set up by these "boundaries"

That is we are interested in the ranges:
#color(white)("XXX")x < -1/2#
#color(white)("XXX")x in (-1/2,1/3)#
#color(white)("XXX")x > 1/3#

We can test each of these ranges by picking any easily evaluated sample value of #x# within each range
and then seeing if that value satisfies the required inequality.

For example:
#color(white)("XXX")#in the range #x < -1/2# we could use #x=-1#
#color(white)("XXX")#in the range #x in (-1/2,1/3)# we could use #x=0#
#color(white)("XXX")#in the range #x > 1/3# we could use #x=+1#

#{:(,"|","in range " x < -1/2,"in range "x in (-1/2,1/3),"in range "x > 1/3), (,"|",underline("sample: " x=-1),underline("sample: "x=0),underline("sample: "x=+1)), (6x^2+x,"|",color(white)("XXX")5,color(white)("XXX")-1,color(white)("XXX")7), (6x^2+x > 1?,"|","True","False","True") :}#