VARIATION: Find all real solutions to the equation: √(variable_93 + 3 - 4√(variable_93 - 1)) + √(variable_93 + 8 - 6√(variable_93 - 1)) = 1. Find the numerical value if exact

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 $(a-b)^2 = a^2 - 2ab + b^2$. Let $a = \sqrt{x-1}$. Then 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, $x+8-6\sqrt{x-1} = (x-1) - 6\sqrt{x-1} + 9 = (\sqrt{x-1})^2 - 2(3)(\sqrt{x-1}) + 3^2 = (\sqrt{x-1} - 3)^2$.

Step 2: Apply the property $\sqrt{a^2} = |a|$

Using the property that $\sqrt{a^2} = |a|$, we can simplify the equation to involve absolute values. This gives us the sum of two absolute value expressions equal to 1.

Step 3: Define a substitution and analyze cases

To simplify the problem, let's substitute $y = \sqrt{x - 1}$. Since $x-1$ must be non-negative for $\sqrt{x-1}$ to be real, we have $x \ge 1$, which implies $y \ge 0$. We need to consider 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 $y < 2$, then $y-2$ is negative and $y-3$ is negative. The equation becomes $-(y-2) - (y-3) = 1$, which simplifies to $-2y+5=1$, giving $y=2$. This contradicts our assumption $y<2$. \\ 2. If $2 \le y \le 3$, then $y-2$ is non-negative and $y-3$ is non-positive. The equation becomes $(y-2) - (y-3) = 1$, which simplifies to $1=1$. This is always true, so all values of $y$ in the interval $[2, 3]$ are solutions. \\ 3. If $y > 3$, then $y-2$ is positive and $y-3$ is positive. The equation becomes $(y-2) + (y-3) = 1$, which simplifies to $2y-5=1$, giving $y=3$. This contradicts our assumption $y>3$. \\ Therefore, the solution for $y$ is $2 \le y \le 3$.

Step 5: Substitute back to find $x$

Now we substitute back $y = \sqrt{x - 1}$ into the inequality $2 \le y \le 3$.

Step 6: Solve for $x$

To solve for $x$, we square all parts of the inequality. This gives $4 \le x - 1 \le 9$. Then, we add 1 to all parts of the inequality to isolate $x$, resulting in $5 \le x \le 10$. This is the range of all real solutions for $x$.

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