How do you solve #-x^{2} - 6x - 9\leq - 4#?

1 Answer
S.S
Sep 22, 2016

#x<=0 and x = 5 or x = 1#

Explanation:

The "greater than or equal to" sign works as an "equals" sign for rearranging. Therefore the equation becomes;

#-x^2 - 6x -5 <= 0 #

Then, we solve the quadratic equation we have formed. Which gives us;
#(-x - 5) or (x + 1) <= 0#

Which goes to

# x = 5 or x = 1 and x<= 0#

Related questions
Impact of this question
487 views around the world
You can reuse this answer
Creative Commons License