Question

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

Answer 1

`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]}

Answer 2

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)