Solve the inequality and graph the solution set on a real number line. Express the solution set in interval notation.

Question

Solve the inequality and graph the solution set on a real number line.  Express the solution set in interval notation.
StartAbsoluteValue x squared plus 4 x minus 22 EndAbsoluteValue greater than 10
Question content area bottom
Part 1
The solution set is 
  
enter your response here.
​(Type your answer in interval notation.  Use integers or fractions for any numbers in the​ expression.)

GPT-4o mini · Answer

To solve the inequality $ |x^2 + 4x - 22| > 10 $, we start by breaking it down into two cases based on the properties of absolute values. ### Case 1: $$ x^2 + 4x - 22 > 10 $$ Rearranging we have: $$ x^2 + 4x - 32 > 0 $$ Next, we can solve the corresponding quadratic equation: $$ x^2 + 4x - 32 = 0 $$ Using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $ a = 1, b = 4, c = -32 $: $$ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm \sqrt{144}}{2} = \frac{-4 \pm 12}{2} $$ Calculating the two roots: 1. $ x = \frac{8}{2} = 4 $ 2. $ x = \frac{-16}{2} = -8 $ Now, we can determine the intervals for which $ x^2 + 4x - 32 > 0 $: The roots divide the number line into intervals: $ (-\infty, -8) $, $ (-8, 4) $, and $ (4, \infty) $. We will test each interval: - For $ x = -9 $ (in $ (-\infty, -8) $): $$ (-9)^2 + 4(-9) - 32 = 81 - 36 - 32 = 13 > 0 \quad \text{(True)} $$ - For $ x = 0 $ (in $ (-8, 4) $): $$ 0^2 + 4(0) - 32 = -32 < 0 \quad \text{(False)} $$ - For $ x = 5 $ (in $ (4, \infty) $): $$ 5^2 + 4(5) - 32 = 25 + 20 - 32 = 13 > 0 \quad \text{(True)} $$ So from Case 1, the solution is: $$ (-\infty, -8) \cup (4, \infty) $$ ### Case 2: $$ -(x^2 + 4x - 22) > 10 $$ This simplifies to: $$ -x^2 - 4x + 22 > 10 $$ Rearranging we get: $$ -x^2 - 4x + 12 > 0 $$ or $$ x^2 + 4x - 12 < 0 $$ Solving the corresponding quadratic equation $ x^2 + 4x - 12 = 0 $: Using the quadratic formula again, where $ a = 1, b = 4, c = -12 $: $$ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm \sqrt{64}}{2} = \frac{-4 \pm 8}{2} $$ Calculating the roots: 1. $ x = \frac{4}{2} = 2 $ 2. $ x = \frac{-12}{2} = -6 $ The roots divide the number line into intervals: $ (-\infty, -6) $, $ (-6, 2) $, and $ (2, \infty) $. We test each interval: - For $ x = -7 $ (in $ (-\infty, -6) $): $$ (-7)^2 + 4(-7) - 12 = 49 - 28 - 12 = 9 > 0 \quad \text{(False)} $$ - For $ x = 0 $ (in $ (-6, 2) $): $$ 0^2 + 4(0) - 12 = -12 < 0 \quad \text{(True)} $$ - For $ x = 3 $ (in $ (2, \infty) $): $$ 3^2 + 4(3) - 12 = 9 + 12 - 12 = 9 > 0 \quad \text{(False)} $$ So from Case 2, the solution is: $$ (-6, 2) $$ ### Final Solution: Combining both cases, the overall solution is: $$ (-\infty, -8) \cup (-6, 2) \cup (4, \infty) $$ In interval notation, the solution set is: $$ \boxed{(-\infty, -8) \cup (-6, 2) \cup (4, \infty)} $$ ### Graphing the solution set: You would represent this on a number line with open circles at points $-8$, $-6$, $2$, and $4$, shading the intervals that are included in the solution set.