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.)
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.)
Answers
GPT-4o mini
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.
### 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.