Question

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

Answer 1

`(-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)`

Answer 2

`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") :}#