Let's solve the equation [tex]\(x^2 - 5 = |x| + 1\)[/tex] and determine which values of [tex]\(x\)[/tex] from the given options satisfy this equation.
Step 1: Understand the equation and possible cases for [tex]\(|x|\)[/tex]:
[tex]\[ x^2 - 5 = |x| + 1 \][/tex]
Here, [tex]\(|x|\)[/tex] is the absolute value of [tex]\(x\)[/tex]. There are two cases to consider based on whether [tex]\(x\)[/tex] is non-negative or negative.
Case 1: [tex]\(x \geq 0\)[/tex]
In this case, [tex]\(|x| = x\)[/tex], and our equation becomes:
[tex]\[ x^2 - 5 = x + 1 \][/tex]
Rearranging terms, we get:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
Factorizing the quadratic equation:
[tex]\[ (x - 3)(x + 2) = 0 \][/tex]
So, the solutions for [tex]\(x\)[/tex] in this case are:
[tex]\[ x = 3 \ \text{or} \ x = -2 \][/tex]
Case 2: [tex]\(x < 0\)[/tex]
In this case, [tex]\(|x| = -x\)[/tex], and our equation becomes:
[tex]\[ x^2 - 5 = -x + 1 \][/tex]
Rearranging terms, we get:
[tex]\[ x^2 + x - 6 = 0 \][/tex]
Factorizing the quadratic equation:
[tex]\[ (x + 3)(x - 2) = 0 \][/tex]
So, the solutions for [tex]\(x\)[/tex] in this case are:
[tex]\[ x = -3 \ \text{or} \ x = 2 \][/tex]
Step 2: Validate the solutions with given options.
Given options:
A. [tex]\(x = -5\)[/tex]
B. [tex]\(x = -4\)[/tex]
C. [tex]\(x = -3\)[/tex]
D. [tex]\(x = -2\)[/tex]
E. [tex]\(x = 1\)[/tex]
F. [tex]\(x = 2\)[/tex]
G. [tex]\(x = 3\)[/tex]
H. [tex]\(x = 4\)[/tex]
From both cases, we have potential solutions: [tex]\(x = 3, -2, -3, 2\)[/tex]. Checking which of these are among the given options:
- [tex]\(x = -3\)[/tex] is option C
- [tex]\(x = 3\)[/tex] is option G
Hence, the valid options that satisfy the equation [tex]\(x^2 - 5 = |x| + 1\)[/tex] are:
[tex]\[ C. \, x = -3 \][/tex]
[tex]\[ G. \, x = 3 \][/tex]
