Determine the interval whereby 6x^2 + 44x + 70 ≥ 0?

1 Answer
Oct 14, 2016

Answer:

#x in (-oo, -5] uu [-7/3, oo)#

Explanation:

#6x^2+44x+70 = (2x+10)(3x+7) >= 0#

The product of two values is positive only if both values are positive or both values are negative. Thus

#2x+10 >= 0 and 3x+7 >= 0#
or
#2x+10 <= 0 and 3x-7 <= 0#

Solving for #x# in each inequality, we find

#x >= -5 and x >= -7/3#
or
#x <= -5 and x <= -7/3#

Because #-7/3 > -5#, we have that #x >= -7/3# implies #x>=-5# already. Similarly, #x<=-5# implies #x<=-7/3#. Thus, we can rewrite the pairs of inequalities as single inequalities:

#x>= -7/3 or x<=-5#

Using interval notation, we can express the set of values which act as solutions for each inequality (note that #in# means "is an element of" or "is in":)

#x >= -7/3 <=> x in [-7/3, oo)#
(#x# is in the interval containing all real numbers from and including #-7/3# to infinity)

#x <= -5 <=> x in (-oo, -5]#
(#x# is in the interval containing all real numbers from negative infinity to, and including, #-5#)

The #uu#, or "union" symbol allows us to treat multiple sets as a single set. If #A# and #B# are sets, then #AuuB# is the set which contains all the elements of #A# and all the elements of #B#. Thus, we can write our entire solution set as a single set by using #uu# to combine the intervals.

#x in (-oo, -5] uu [-7/3, oo)#