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Solution of the equation $\sqrt {x + 3 - 4\sqrt {x - 1} } + \sqrt {x + 8 - 6\sqrt {x - 1} } = 1$ is
$x \in \left[ {4,9} \right]$
$x \in \left[ {3,8} \right]$
$x \in \left[ {5,10} \right]$
$x \in \left[ {4,7} \right]$
Solution
Let $u=x-1$. We can then rewrite the left hand side of the equation as
$\sqrt{u+4-4 \sqrt{u}}+\sqrt{u+9-6 \sqrt{u}}$
$=\sqrt{(\sqrt{u}-2)^{2}}+\sqrt{(\sqrt{u}-3)^{2}}$
$=|\sqrt{u}-2|+|\sqrt{u}-3|$
Note the presence of $\sqrt{u}$ in the equation and that we are only looking for real values, so we have the restriction $u \geq 0$. With that, we will now consider all remaining cases:
Case $1: 0 \leq u \leq 4$
$|\sqrt{u}-2|+|\sqrt{u}-3|=1$
$\Rightarrow 2-\sqrt{u}+3-\sqrt{2}=1$
$\Rightarrow-2 \sqrt{u}=-4$
$\Rightarrow \sqrt{u}=2$
$\Rightarrow u=4$
Thus $u=4$ is the only solution in the interval [0,4]
Case $2: 4 \leq u \leq 9$
$|\sqrt{u}-2|+|\sqrt{u}-3|=1$
$\Rightarrow \sqrt{u}-2+3-\sqrt{u}=1$
$\Rightarrow 1=1$
As this is a tautology, every value in [4,9] is a solution.
Case $3: u \geq 9$
$|\sqrt{u}-2|+|\sqrt{u}-3|=1$
$\Rightarrow \sqrt{u}-2+\sqrt{u}-3=1$
$\Rightarrow 2 \sqrt{u}=6$
$\Rightarrow \sqrt{u}=3$
$\Rightarrow u=9$
Thus $u=9$ is the only solution in the interval $[9, \infty)$
Taken together, we have [4,9] as the solution set for real values of $u$. Substituting in $x=u+1,$ we arrive at the final solution set $x \in[5,10]$