Question

x^2 < x+12 in interval notation. Im not sure how to work with the x^2

Answer 1

x^2- x - 12 < 0

so what you are considering is where the parabola
y <b>=</b> x^2 - x - 12 is below the x-axis

x^2 - x - 12
= (x-4)(x+3)
so the x -intercepts are -3 and 4 and the parabola opens up, and so

for x^2 < x+12
-3 < x < 4

I will let you put that in the "interval" notation you learned.
(In the olden days we used the method I just gave you. )