To solve the equation [tex]\(\left|x^2-2x-1\right|=7\)[/tex], we need to handle the absolute value properly. The absolute value equation [tex]\(\left|A\right|=B\)[/tex] can be split into two separate equations:
1. [tex]\(A = B\)[/tex]
2. [tex]\(A = -B\)[/tex]
In our case, [tex]\(A = x^2-2x-1\)[/tex] and [tex]\(B = 7\)[/tex]. Thus, we need to solve the following two equations:
1. [tex]\(x^2 - 2x - 1 = 7\)[/tex]
2. [tex]\(x^2 - 2x - 1 = -7\)[/tex]
Let's tackle each equation separately:
### Solving [tex]\(x^2 - 2x - 1 = 7\)[/tex]
1. Rearrange the equation to standard quadratic form:
[tex]\[
x^2 - 2x - 1 - 7 = 0
\][/tex]
[tex]\[
x^2 - 2x - 8 = 0
\][/tex]
2. Factor the quadratic equation:
[tex]\[
(x - 4)(x + 2) = 0
\][/tex]
3. Solve for [tex]\(x\)[/tex]:
[tex]\[
x - 4 = 0 \implies x = 4
\][/tex]
[tex]\[
x + 2 = 0 \implies x = -2
\][/tex]
So the solutions from this equation are [tex]\(x = 4\)[/tex] and [tex]\(x = -2\)[/tex].
### Solving [tex]\(x^2 - 2x - 1 = -7\)[/tex]
1. Rearrange the equation to standard quadratic form:
[tex]\[
x^2 - 2x - 1 + 7 = 0
\][/tex]
[tex]\[
x^2 - 2x + 6 = 0
\][/tex]
2. Since this is a quadratic equation, we use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = -2\)[/tex], and [tex]\(c = 6\)[/tex]:
[tex]\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{2 \pm \sqrt{4 - 24}}{2}
\][/tex]
[tex]\[
x = \frac{2 \pm \sqrt{-20}}{2}
\][/tex]
[tex]\[
x = \frac{2 \pm 2i\sqrt{5}}{2}
\][/tex]
[tex]\[
x = 1 \pm i\sqrt{5}
\][/tex]
The solutions from this equation are complex numbers: [tex]\(x = 1 + i\sqrt{5}\)[/tex] and [tex]\(x = 1 - i\sqrt{5}\)[/tex].
### Final Solution
Combining all real and complex solutions, the complete set of solutions to the equation [tex]\(\left|x^2-2 x-1\right|=7\)[/tex] is:
[tex]\[
x = \{ 4, -2, 1 + i\sqrt{5}, 1 - i\sqrt{5} \}
\][/tex]
Thus, the equation has two real solutions, [tex]\(x = 4\)[/tex] and [tex]\(x = -2\)[/tex], and two complex solutions, [tex]\(x = 1 + i\sqrt{5}\)[/tex] and [tex]\(x = 1 - i\sqrt{5}\)[/tex].
