#3x^2-5x>8#
Subtract #8# from both sides:
#3x^2-5x-8>0#
We now solve the equation:
#3x^2-5x-8=0#
This will give us the boundary values of #x#.
Factoring:
#(3x-8)(x+1)=0=> x=-1 and x= 8/3#
METHOD 1
From:
#3x^2-5x-8>0#
Notice that the coefficient of #x^2>0#, this means the parabola is in this form:
#uuu#
For values greater than #0# we will be above the #x# axis.
Since the roots to the equation are on the #x# axis we can see that for positive #y# values we will be to the left of #x=-1# and to the right of #x=8/3#.
Since this is a #># and not a #>= # inequality #x=-1# and #x=8/3# are not included points. So the solution in interval notation we be a union of intervals:
#(-oo,-1)uu(8/3,oo)#
METHOD 2
Using the factors of #3x^2-5x-8>0# we found earlier, we can see that:
#(3x-8)(x+1)#
Will be positive i.e. #>0# if both brackets are positive or both brackets are negative. We can use a table to check this using the following inequalities:
#x<-1# , #color(white)(88)-1< x< 8/3# , #color(white)(88) 8/3 < x#
You can see from the table that if we assign #x# a value satisfying the inequality in each column. it is only for the inequalities #x> -1# and #8/3 < x# that the product of the brackets is positive.
So our solution in interval notation is:
#(-oo,-1)uu(8/3 , oo)color(white(88)# as before.
Graph
graph{y<3x^2-5x-8 [-32.48, 32.46, -16.24, 16.25]}
