How do you solve the quadratic inequality #x^2-9x +18 >0#?

2 Answers
Aug 6, 2015

Answer:

#x^2-9x+18 = (x-3)(x-6)# so #x^2-9x+18 = 0# when #x=3# and #x=6#. Since the coefficient of the #x^2# term is positive, #x^2-9x+18 > 0# when #x in (-oo, 3) uu (6, oo)#.

Explanation:

#x^2-9x+18 = (x-3)(x-6)# has zeros #x=3# and #x=6#.

When #x < 3# both #(x-3) < 0# and #(x-6) < 0#, so #(x-3)(x-6) > 0#

When #3 <= x <= 6# we have #(x-3) >= 0# and #(x-6) <= 0#, so #(x-3)(x-6) <= 0#.

When #x > 6#, both #(x-3) > 0# and #(x-6) > 0#, so #(x-3)(x-6) > 0#

Putting these together:

#x^2-9x+18 = (x-3)(x-6) > 0#

for #x in (-oo, 3) uu (6, oo)#

graph{x^2-9x+18 [-5.71, 14.29, -3.68, 6.32]}

Aug 16, 2015

Answer:

Solve f(x) = x^2 - 9x + 18 > 0

Ans: (-infinity, 3) and (5, infinity)

Explanation:

First, solve f(x) = x^2 - 9x + 18 = 0.
Roots have same sign. Factor pairs of (18) --> (2, 9)(3, 6). This sum is 9 = -b. Then the 2 real roots are: 3 and 6.
Use the algebraic method to solve f(x) > 0. Between the 2 real roots
(3) and (5), f(x) < 0 as opposite to the sign of a = 1. f(x) is positive (> 0) outside the interval (3, 5).
Answer by open intervals: (-infinity, 3) and (5, infinity)