How do you solve and write the following in interval notation: #x^2 < 3x + 18#?

2 Answers
Mar 28, 2017

Answer:

#(-3,6)#

Explanation:

Rearrange inequality to have zero on the right side.

#rArrx^2-3x-18<0#

Factorise the quadratic on the left side.

#rArr(x-6)(x+3)<0#

For the left side to be negative, that is < 0 , then.

#x-6>0color(red)" and " x+3<0#

#rArrx>6color(red)" and " x<-3larrcolor(blue)" impossible"#

#color(magenta)" OR"#

#x-6<0color(red)" and " x+3>0#

#rArrx<6color(red)" and " x> -3larrcolor(blue)" solution"#

#rArrx in(-3,6)#

Mar 28, 2017

Answer:

#x in (-3,6)#

Explanation:

Given
#color(white)("XXX")x^2 < 3x+18#

Re-write as:
#color(white)("XXX")x^2-3x-18 < 0#

#color(white)("XXX")(x-6)(x+3) < 0#

Note that #x!=0# since that would make the left side equal to #0#.

The left side is a quadratic dividing the domain of #x# (after excluding #{-3,6}# into 3 subsets:

#{: (," | ",underline(x < -3),color(white)("X"),underline(x in (-3,6)),color(white)("X"),ul(x > 6)), (ul"Sample "," | ",ul(x=-4),,ul(x=0),,ul(x=7)), (x^2," | ",16,,0,,49), (ul(3x+18)," | ",ul(6),,ul(18),,ul(39)), (x^2 < 3x+18?," | ","False",,"True",,"False") :}#