### Instruction: Find all solutions to
\[\sqrt{x + 3 - 4 \sqrt{x - 1}} + \sqrt{x + 8 - 6 \sqrt{x - 1}} = 1.\].### Response: This problem looks challenging, but I notice that the expressions inside the square roots are both of the form $a - b \sqrt{x - 1}$, where $a$ and $b$ are constants.
Maybe I can use a substitution to simplify the equation.
Let me try letting $y = \sqrt{x - 1}$, so that $x = y^2 + 1$.
Then the equation becomes \[\sqrt{y^2 + 4 - 4y} + \sqrt{y^2 + 9 - 6y} = 1.\]
This looks a bit better, but I still have square roots to deal with.
I wonder if I can factor the expressions inside the square roots.
For the first one, I need two numbers that multiply to $4$ and add to $-4$.
Those are $-2$ and $-2$.
So I can write \[\sqrt{y^2 + 4 - 4y} = \sqrt{(y - 2)^2} = |y - 2|.\]
For the second one, I need two numbers that multiply to $9$ and add to $-6$.
Those are $-3$ and $-3$.
So I can write \[\sqrt{y^2 + 9 - 6y} = \sqrt{(y - 3)^2} = |y - 3|.\]
Now the equation becomes \[|y - 2| + |y - 3| = 1.\]
This is much simpler to work with.
I know that the absolute value function changes its sign depending on whether the argument is positive or negative.
So I can split the equation into three cases: $y < 2$, $2 \leq y \leq 3$, and $y > 3$.
For the first case, $y < 2$, both $y - 2$ and $y - 3$ are negative, so the equation becomes \[-(y - 2) - (y - 3) = 1.\]
Simplifying, I get \[-2y + 5 = 1.\]
Solving for $y$, I get \[y = 2.\]
But wait, this is not a valid solution, because it contradicts the assumption that $y < 2$.
So there are no solutions in this case.
For the second case, $2 \leq y \leq 3$, $y - 2$ is nonnegative and $y - 3$ is nonpositive, so the equation becomes \[y - 2 - (y - 3) = 1.\]
Simplifying, I get \[1 = 1.\]
This is true for any value of $y$ in the interval $[2, 3]$.
So any value of $y$ in this interval is a solution.
For the third case, $y > 3$, both $y - 2$ and $y - 3$ are positive, so the equation becomes \[y - 2 + y - 3 = 1.\]
Simplifying, I get \[2y - 5 = 1.\]
Solving for $y$, I get \[y = 3.\]
But again, this is not a valid solution, because it contradicts the assumption that $y > 3$.
So there are no solutions in this case either.
Therefore, the only solutions are the values of $y$ in the interval $[2, 3]$.
To find the corresponding values of $x$, I need to plug in $y = \sqrt{x - 1}$.
This means that $x - 1$ must be in the interval $[4, 9]$, since those are the squares of $2$ and $3$.
Adding $1$ to both sides, I get that $x$ must be in the interval $[5, 10]$.
So the final answer is that the solutions are all the values of $x$ in the interval $[5, 10]$.