Quadratic Inequality 3

[mathdad]

Solve the inequality.

x^2 + x > 12

Solution:

x^2 + x - 12 > 0

x^2 + x - 12 = 0

(x - 3)(x + 4) = 0

x = 3, x = - 4

Pick a number between 3 and -4.

Let x = 0

x^2 + x - 12 > 0

(0)^2 + 0 - 12 > 0

-12 > 0...false statement.

I conclude that the solution for the given quadratic inequality is found in the following intervals:

(-infinity, -4) (3, infinity)

Correct?

 

[ksdhart2]

Strictly speaking, you need a union symbol in there:

\(\displaystyle (-\infty, 4) \cup (3, \infty)\)

But otherwise, your solution is correct.

 

[mathdad]

Strictly speaking, you need a union symbol in there:

\(\displaystyle (-\infty, 4) \cup (3, \infty)\)

But otherwise, your solution is correct.

I'll remember to place a union symbol U next time.