How do you solve #(x + 13)(x - 19)>=0#?

2 Answers
Aug 10, 2015

Answer:

Answer: #x in (-oo;-13> uu <19;+oo)#

Explanation:

From tjhe inequality you can see, that your function has 2 zeros: #x_1=-13# and #x_2=19# and the function takes positive values when #x# goes to #+oo# and #-oo#

graph{x^2-6x-247 [-40, 40, -20, 20]}

so you can write the solution: #x in (-oo;-13> uu <19;+oo)#

Aug 16, 2015

Answer:

Solve # (x + 13)(x - 19) >= 0#

Ans: (-infinity, -13] and [19, infinity)

Explanation:

The 2 x-intercepts (real roots) are x = -13 and x = 19.
Use the algebraic method to solve the inequality #f(x) >= 0#. Between the 2 real roots f(x) < 0 as opposite to the side of a = 1.. Outside the interval (-13, 19), #f(x) >= 0.#
Answer by half closed intervals: (-infinity, -13] and [19, infinity).
The critical points (-13) and (19) are included in the solution set.