How do you solve #X^2 +2X > 3#?

1 Answer
Mar 25, 2018

Answer:

The solution set is #x<-3 or x>1#.

Explanation:

#x^2+2x>3#

#x^2+2x-3>0#

#(x+3)(x-1)>0#

Since we're trying to find when the function is greater than #0#, and the parabola opens upwards, that means the inequality is true whenever #x# is NOT between the zeroes.

This makes more sense if you look at the graph:

graph{x^2+2x-3 [-11, 9, -5, 5]}

Since the zeroes are #-3# and #1#, the inequality is less than #-3# or greater #1#, or:

#x<-3 or x>1#

That's the solution. Hope this helped!