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

Aug 6, 2015

${x}^{2} - 9 x + 18 = \left(x - 3\right) \left(x - 6\right)$ so ${x}^{2} - 9 x + 18 = 0$ when $x = 3$ and $x = 6$. Since the coefficient of the ${x}^{2}$ term is positive, ${x}^{2} - 9 x + 18 > 0$ when $x \in \left(- \infty , 3\right) \cup \left(6 , \infty\right)$.

Explanation:

${x}^{2} - 9 x + 18 = \left(x - 3\right) \left(x - 6\right)$ has zeros $x = 3$ and $x = 6$.

When $x < 3$ both $\left(x - 3\right) < 0$ and $\left(x - 6\right) < 0$, so $\left(x - 3\right) \left(x - 6\right) > 0$

When $3 \le x \le 6$ we have $\left(x - 3\right) \ge 0$ and $\left(x - 6\right) \le 0$, so $\left(x - 3\right) \left(x - 6\right) \le 0$.

When $x > 6$, both $\left(x - 3\right) > 0$ and $\left(x - 6\right) > 0$, so $\left(x - 3\right) \left(x - 6\right) > 0$

Putting these together:

${x}^{2} - 9 x + 18 = \left(x - 3\right) \left(x - 6\right) > 0$

for $x \in \left(- \infty , 3\right) \cup \left(6 , \infty\right)$

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

Aug 16, 2015

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)