# How do you solve x^2 + x - 12 > 0?

Jun 19, 2015

Cut the number line into pieces (partition the number line) and test on each piece to see where the inequality is true. The solution set is $\left(- \infty , - 4\right) \cup \left(3 , \infty\right)$

#### Explanation:

We'll use the general method (there are other method that only work for quadratics, but I'll use the more general method.)

It is not necessary, but later steps are easier if we first make sure that we have $0$ on one side. (Any constant will be ok, but $0$ is easiest.)

Partition the number line by finding zeros of the expression and points at which it is not defined.
${x}^{2} + x - 12$ is defined for all $x$ so we only need to find its zeros

Solve ${x}^{2} + x - 12 = 0$ by factoring if possible. (By formula or completing the square if you can't factor it.)

$\left(x - 3\right) \left(x + 4\right) = 0$, so the partition numbers (aka key numbers) are: $3 \mathmr{and} - 4$

We cut the number line into pieces (intervals): $\left(- \infty , - 4\right) , \left(- 4 , 3\right) , \left(3 , \infty\right)$

We now test on each piece to see whether
${x}^{2} + x - 12$ is positive ($> 0$) or negative ($< 0$)

Since we've already factored the expression, we can check using the factored form:

On $\left(- \infty , - 4\right)$ both factors $\left(x - 3\right)$ and $\left(x + 4\right)$ are negative (try $x = - 10$) , so the product is positive. The inequality asks us to make the expression positive, so the interval $\left(- \infty , - 4\right)$ is part of the solution set.

On $\left(- 4 , 3\right)$, we have $x - 3$ is negative (try $x = 0$ and $x + 4$ is positive. The product is negative so this is not part of the solution set.

On $\left(3 , \infty\right)$, both factors are positive (try $x = 10$), so the product is positive and this interval is part of the solution.

The solution set is $\left(- \infty , - 4\right) \cup \left(3 , \infty\right)$

Note
The endpoints, $- 4 \mathmr{and} 3$ are not included because the inequality is strict and the expression is equal to $0$ at the endpoints.

Note 2 (for the future)

The inequalities:

$\frac{x + 4}{x - 3} > 0$ and $\frac{x - 3}{x + 4} > 0$ are solved (almost) exactly the same way and they have exactly the same solution set.
(Almost) because we have "quotient" instead of "product".