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

1 Answer
Oct 14, 2016

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)