Questions: Solve the inequality algebraically and write any solution in interval notation. (If there is no solution, enter NO SOLUTION.) x^2+7 x>-12

Transcript text: Solve the inequality algebraically and write any solution in interval notation. (If there is no solution, enter NO SOLUTION.) \[ x^{2}+7 x>-12 \] $\square$

Solution

Solution Steps

Step 1: Rewrite the inequality in standard form

Move all terms to one side of the inequality to set it to zero: \[ x^{2} + 7x + 12 > 0 \]

Step 2: Factor the quadratic expression

Factor the quadratic expression $ x^{2} + 7x + 12 $: \[ (x + 3)(x + 4) > 0 \]

Step 3: Find the critical points

Set each factor equal to zero to find the critical points: \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]

Step 4: Determine the intervals to test

The critical points divide the number line into three intervals:

$ x < -4 $
$ -4 < x < -3 $
$ x > -3 $

Step 5: Test each interval

Choose a test point from each interval and substitute it into the inequality $ (x + 3)(x + 4) > 0 $:

For $ x < -4 $, choose $ x = -5 $: \[ (-5 + 3)(-5 + 4) = (-2)(-1) = 2 > 0 \quad \text{(True)} \]
For $ -4 < x < -3 $, choose $ x = -3.5 $: \[ (-3.5 + 3)(-3.5 + 4) = (-0.5)(0.5) = -0.25 < 0 \quad \text{(False)} \]
For $ x > -3 $, choose $ x = 0 $: \[ (0 + 3)(0 + 4) = (3)(4) = 12 > 0 \quad \text{(True)} \]

Step 6: Write the solution in interval notation

The inequality holds true for $ x < -4 $ and $ x > -3 $. Therefore, the solution in interval notation is: \[ (-\infty, -4) \cup (-3, \infty) \]