Find all real solutions to the equation: √(x + 3 - 4√ $(x - 1)$) + √(x + 8 - 6√ $(x - 1)$) = 1.

Step 1: Simplify the terms under the square roots

We observe that the expressions under the square roots resemble the expansion of a perfect square $(a - b)^2 = a^2 - 2ab + b^2$. By letting $a = \sqrt{x - 1}$, we can rewrite the terms. For the first term, we have $x + 3 - 4\sqrt{x - 1} = (x - 1) - 4\sqrt{x - 1} + 4 = (\sqrt{x - 1})^2 - 2(2)(\sqrt{x - 1}) + 2^2 = (\sqrt{x - 1} - 2)^2$. Similarly, for the second term, $x + 8 - 6\sqrt{x - 1} = (x - 1) - 6\sqrt{x - 1} + 9 = (\sqrt{x - 1})^2 - 2(3)(\sqrt{x - 1}) + 3^2 = (\sqrt{x - 1} - 3)^2$. Taking the square root of a square results in the absolute value.

Step 2: Substitute simplified terms into the equation

Now we substitute the simplified absolute value expressions back into the original equation. This transforms the problem into solving an equation involving absolute values.

Step 3: Define a substitution and analyze cases

To simplify the equation further, let's introduce a substitution: $y = \sqrt{x - 1}$. Since the square root must be non-negative, $y \ge 0$. Also, for $\sqrt{x-1}$ to be defined, $x-1 \ge 0$, which means $x \ge 1$. We now need to solve the absolute value equation $|y - 2| + |y - 3| = 1$. We will analyze this equation by considering different cases for $y$ based on the critical points $y=2$ and $y=3$.

Step 4: Solve for y in different cases

We analyze three cases for $y$: \\ 1. If $0 \le y < 2$: Both $(y-2)$ and $(y-3)$ are negative, so $|y-2| = -(y-2) = 2-y$ and $|y-3| = -(y-3) = 3-y$. The equation becomes $(2-y) + (3-y) = 1$, which simplifies to $5 - 2y = 1$, giving $2y = 4$, so $y = 2$. This contradicts our assumption that $y < 2$, so there are no solutions in this case. \\ 2. If $2 \le y < 3$: $(y-2)$ is non-negative, so $|y-2| = y-2$. $(y-3)$ is negative, so $|y-3| = -(y-3) = 3-y$. The equation becomes $(y-2) + (3-y) = 1$, which simplifies to $1 = 1$. This statement is true for all $y$ in the interval $2 \le y < 3$. \\ 3. If $y \ge 3$: Both $(y-2)$ and $(y-3)$ are non-negative, so $|y-2| = y-2$ and $|y-3| = y-3$. The equation becomes $(y-2) + (y-3) = 1$, which simplifies to $2y - 5 = 1$, giving $2y = 6$, so $y = 3$. This is consistent with our assumption $y \ge 3$. \\ Combining these cases, the solutions for $y$ are $2 \le y \le 3$.

Step 5: Substitute back to find x

Now we substitute back $y = \sqrt{x - 1}$ into the inequality $2 \le y \le 3$. This gives us $2 \le \sqrt{x - 1} \le 3$. To solve for $x$, we square all parts of the inequality, which is valid since all terms are non-negative. This yields $4 \le x - 1 \le 9$. Finally, we add 1 to all parts of the inequality to isolate $x$, resulting in $5 \le x \le 10$. All values of $x$ in this interval satisfy the original equation.

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