Answer :
To solve the equation [tex]\(\left|x^2 + 2x - 36\right| = 12\)[/tex], we need to consider the definition of absolute value. Absolute value equations split into two separate cases, depending on whether the expression inside the absolute value is positive or negative.
### Step-by-Step Solution:
1. Consider the two possible cases for the equation:
- Case 1: [tex]\(x^2 + 2x - 36 = 12\)[/tex]
- Case 2: [tex]\(x^2 + 2x - 36 = -12\)[/tex]
2. Solve each equation separately:
Case 1: [tex]\(x^2 + 2x - 36 = 12\)[/tex]
- Subtract 12 from both sides to set the equation to zero:
[tex]\[ x^2 + 2x - 36 - 12 = 0 \][/tex]
This simplifies to:
[tex]\[ x^2 + 2x - 48 = 0 \][/tex]
- Solve the quadratic equation:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -48\)[/tex]:
[tex]\[ x = \frac{{-2 \pm \sqrt{{2^2 - 4 \cdot 1 \cdot (-48)}}}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{{-2 \pm \sqrt{{4 + 192}}}}{2} \][/tex]
[tex]\[ x = \frac{{-2 \pm \sqrt{196}}}{2} \][/tex]
[tex]\[ x = \frac{{-2 \pm 14}}{2} \][/tex]
This gives the solutions:
[tex]\[ x = \frac{12}{2} = 6 \quad \text{and} \quad x = \frac{-16}{2} = -8 \][/tex]
Case 2: [tex]\(x^2 + 2x - 36 = -12\)[/tex]
- Add 12 to both sides to set the equation to zero:
[tex]\[ x^2 + 2x - 36 + 12 = 0 \][/tex]
This simplifies to:
[tex]\[ x^2 + 2x - 24 = 0 \][/tex]
- Solve the quadratic equation:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -24\)[/tex]:
[tex]\[ x = \frac{{-2 \pm \sqrt{{2^2 - 4 \cdot 1 \cdot (-24)}}}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{{-2 \pm \sqrt{{4 + 96}}}}{2} \][/tex]
[tex]\[ x = \frac{{-2 \pm \sqrt{100}}}{2} \][/tex]
[tex]\[ x = \frac{{-2 \pm 10}}{2} \][/tex]
This gives the solutions:
[tex]\[ x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-12}{2} = -6 \][/tex]
3. Combine and sort the solutions:
The solutions from both cases are:
[tex]\(x = 6\)[/tex], [tex]\(x = -8\)[/tex], [tex]\(x = 4\)[/tex], and [tex]\(x = -6\)[/tex].
Sorted in ascending order, the solutions are:
[tex]\[ x = -8, -6, 4, 6 \][/tex]
### Final Answer:
[tex]\[ x = -8.0, -6.0, 4.0, 6.0 \][/tex]
### Step-by-Step Solution:
1. Consider the two possible cases for the equation:
- Case 1: [tex]\(x^2 + 2x - 36 = 12\)[/tex]
- Case 2: [tex]\(x^2 + 2x - 36 = -12\)[/tex]
2. Solve each equation separately:
Case 1: [tex]\(x^2 + 2x - 36 = 12\)[/tex]
- Subtract 12 from both sides to set the equation to zero:
[tex]\[ x^2 + 2x - 36 - 12 = 0 \][/tex]
This simplifies to:
[tex]\[ x^2 + 2x - 48 = 0 \][/tex]
- Solve the quadratic equation:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -48\)[/tex]:
[tex]\[ x = \frac{{-2 \pm \sqrt{{2^2 - 4 \cdot 1 \cdot (-48)}}}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{{-2 \pm \sqrt{{4 + 192}}}}{2} \][/tex]
[tex]\[ x = \frac{{-2 \pm \sqrt{196}}}{2} \][/tex]
[tex]\[ x = \frac{{-2 \pm 14}}{2} \][/tex]
This gives the solutions:
[tex]\[ x = \frac{12}{2} = 6 \quad \text{and} \quad x = \frac{-16}{2} = -8 \][/tex]
Case 2: [tex]\(x^2 + 2x - 36 = -12\)[/tex]
- Add 12 to both sides to set the equation to zero:
[tex]\[ x^2 + 2x - 36 + 12 = 0 \][/tex]
This simplifies to:
[tex]\[ x^2 + 2x - 24 = 0 \][/tex]
- Solve the quadratic equation:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -24\)[/tex]:
[tex]\[ x = \frac{{-2 \pm \sqrt{{2^2 - 4 \cdot 1 \cdot (-24)}}}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{{-2 \pm \sqrt{{4 + 96}}}}{2} \][/tex]
[tex]\[ x = \frac{{-2 \pm \sqrt{100}}}{2} \][/tex]
[tex]\[ x = \frac{{-2 \pm 10}}{2} \][/tex]
This gives the solutions:
[tex]\[ x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-12}{2} = -6 \][/tex]
3. Combine and sort the solutions:
The solutions from both cases are:
[tex]\(x = 6\)[/tex], [tex]\(x = -8\)[/tex], [tex]\(x = 4\)[/tex], and [tex]\(x = -6\)[/tex].
Sorted in ascending order, the solutions are:
[tex]\[ x = -8, -6, 4, 6 \][/tex]
### Final Answer:
[tex]\[ x = -8.0, -6.0, 4.0, 6.0 \][/tex]