How do you solve #x^2 + 7x + 6 <= 6#?

1 Answer
Oct 5, 2016

Answer:

Solution is #-7<= x <= 0#

Explanation:

As #x^2+7x+6<=6#, we have #x^2+7x+cancel6<=cancel6# or

#x^2+7x<=0# or #x(x+7)<=0#

If product of #x# and #x+7# is negative (for the time we are ignoring equality sign - note that for equality we just add #x=0# and #x=-7# to the solution),

either #x>0# and #x+7<0# i.e. #x<-7# - but #x>0# and #x<-7# together is just not possible.

or #x<0# and #x+7>0#, i.e. #x>-7# - which is possible if #x# lies between #-7# and #0# i.e. #-7< x < 0#.

Now including the equality sign

Solution is #-7<= x <= 0#.

You may also see from the graph that #x^2+7x# is negative in the interval #-7<= x <= 0#.
graph{x(x+7) [-10, 5, -20, 20]}