How do you solve the following Quadratic Inequality #x^2+2x-15<0#?

1 Answer
Zor Shekhtman
Sep 11, 2015

The solution is
#-5 < x < 3#

Explanation:

There are more than one ways to solve this inequality.

Solution #A)#
Since #x^2+2x-15 = (x+5)(x-3)#, we can suggest the following reasoning.
The product of two real numbers can be negative if one of them is positive and another is negative.
Therefore, we have two solutions:
#A_1)# #x+5 > 0# AND #x-3 < 0#
#A_2)# #x+5 < 0# AND #x-3 > 0#

The case #A_1# defines #x# as
#x > -5# AND #x < 3#, which defines an interval for #x#:
#-5 < x < 3#

The case #A_2# defines #x# as
#x < -5# AND #x > 3#, which is impossible.

So, the solution is
#-5 < x < 3#

Solution #B)#
As we know, the graph of the quadratic polynomial on the left is parabola. Since the coefficient at #x^2# is positive, this parabola directs its endpoints upward.
Therefore, the only way it can be negative is in-between its roots, where it's equal to zero.
In other words, the solutions to a inequality is the area between solutions to equality
#x^2+2x-15=0#

Obvious solutions are #x=3# and #x=-5#.
So, the solutions are #-5 < x < 2#

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