How do you solve #n^2+2n-24>0#?

1 Answer
Apr 21, 2017

Answer:

#-4< n>6#

Explanation:

#n^2+color(red)(2)ncolor(orange)(-24)> 0#

We need to factor this quadratic

We need to find two numbers that add to #2# and multiply to #-24#
#+2#
x#24#
..........
#color(black)(1)color(white)(.)color(black)(24)#
#color(black)(2)color(white)(.)color(black)(12)#
#color(black)(3)color(white)(.)color(black)(8)#
#color(green)(4)color(white)(.)color(green)(6)#

#4# and #6# can be combined to form #2#, but we'll need to subtract them.

We are looking for one positive number and one negative number. We can tell because #-24# can only be the product of one positive and negative number.

Now we just need to see if #-4+6# gives us #color(red)(2)# or if it's #-6+4#:
#-4+6=2#
#-6+4=-2#

So, our numbers are #-4# and #6#, which gives us #(n-4)(n+6)> 0#

We just set each factor to zero and solve:
Case 1
#n-4> 0#

#n>4#

Case 2
#n+6> 0#

#n>6#