How do you solve 6x^2+x>16x2+x>1?

1 Answer
Sep 2, 2016

color(green)(x < -1/2)x<12 or color(green)(x > 1/3)x>13

Explanation:

6x^2+x > 16x2+x>1 is equivalent to 6x^2+x-1 > 06x2+x1>0

Factoring 6x^2+x-16x2+x1
we have (3x-1)(2x+1)(3x1)(2x+1)
which will be equal to zero when x=1/3x=13 or x=-1/2x=12

We not interested in when 6x^2+x-16x2+x1 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 < -1/2XXXx<12
color(white)("XXX")x in (-1/2,1/3)XXXx(12,13)
color(white)("XXX")x > 1/3XXXx>13

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

For example:
color(white)("XXX")XXXin the range x < -1/2x<12 we could use x=-1x=1
color(white)("XXX")XXXin the range x in (-1/2,1/3)x(12,13) we could use x=0x=0
color(white)("XXX")XXXin the range x > 1/3x>13 we could use x=+1x=+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") :}