To solve the equation [tex]\( x + \sqrt{x^2 + \sqrt{x^3 + 1}} = 1 \)[/tex], let's analyze it step-by-step.
First, let's isolate the square root term:
[tex]\[
\sqrt{x^2 + \sqrt{x^3 + 1}} = 1 - x
\][/tex]
Next, we need to ensure that the expression under the square root is non-negative, which means [tex]\( 1 - x \geq 0 \)[/tex] or [tex]\( x \leq 1 \)[/tex].
Now, square both sides to eliminate the square root:
[tex]\[
x^2 + \sqrt{x^3 + 1} = (1 - x)^2
\][/tex]
Expanding the right-hand side:
[tex]\[
x^2 + \sqrt{x^3 + 1} = 1 - 2x + x^2
\][/tex]
Next, simplify the equation by subtracting [tex]\( x^2 \)[/tex] from both sides:
[tex]\[
\sqrt{x^3 + 1} = 1 - 2x
\][/tex]
Again, to eliminate the square root, square both sides:
[tex]\[
x^3 + 1 = (1 - 2x)^2
\][/tex]
Expanding the right-hand side:
[tex]\[
x^3 + 1 = 1 - 4x + 4x^2
\][/tex]
Rearranging terms to form a polynomial equation:
[tex]\[
x^3 - 4x^2 + 4x + 1 - 1 = 0
\][/tex]
[tex]\[
x^3 - 4x^2 + 4x = 0
\][/tex]
Factor out [tex]\( x \)[/tex]:
[tex]\[
x(x^2 - 4x + 4) = 0
\][/tex]
This gives us:
[tex]\[
x = 0 \quad \text{or} \quad x^2 - 4x + 4 = 0
\][/tex]
Solve the quadratic equation [tex]\( x^2 - 4x + 4 = 0 \)[/tex]:
[tex]\[
(x - 2)^2 = 0
\][/tex]
This gives us:
[tex]\[
x = 2
\][/tex]
However, we need to verify if these solutions satisfy the original equation. Let's substitute [tex]\( x = 0 \)[/tex] into the original equation:
[tex]\[
0 + \sqrt{0^2 + \sqrt{0^3 + 1}} = 1
\][/tex]
[tex]\[
\sqrt{1} = 1
\][/tex]
Thus, [tex]\( x = 0 \)[/tex] is indeed a solution.
For [tex]\( x = 2 \)[/tex]:
[tex]\[
2 + \sqrt{2^2 + \sqrt{2^3 + 1}} \neq 1
\][/tex]
It is easy to see that substituting [tex]\( x = 2 \)[/tex] would yield a value much greater than 1.
Therefore, the final solution to the equation [tex]\( x + \sqrt{x^2 + \sqrt{x^3 + 1}} = 1 \)[/tex] is:
[tex]\[
x = 0
\][/tex]
