How do you solve the inequality #x^2 - 9x> -18#?

2 Answers
Apr 23, 2018

Answer:

The solutions are #x in (-oo,3) uu(6, +oo)#

Explanation:

The inequality is

#x^2-9x > -18#

#x^2-9x +18 > 0#

#(x-3)(x-6) >0#

Let #f(x)=(x-3)(x-6)#

Let's build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##3##color(white)(aaaaa)##6##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##x-3##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-6##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x) >0# when #x in (-oo,3) uu(6, +oo)#

graph{x^2-9x+18 [-4.29, 15.71, -3.96, 6.04]}

Apr 23, 2018

Answer:

(-inf., 3) and (6, +inf.)

Explanation:

#x^2 - 9x > - 18$ #f(x) = x^2 - 9x + 18 > 0#
Find 2 real roots, that have same sign(ac > 0), knowing the sum (- b = 9) and the product (c = 18). They are 6 and 3.
The graph of f(x) is an upward parabola (a > 0).
Inside the interval (3, 6) --> f(x) < 0 as the graph is below the x-axis.
Outside the interval (3, 6) --> f (x )> 0. Therefor, the solution set is the 2 open intervals:
(- inf., 3) and (6, +inf.)