The solution set of the equation sqrt{x + 3 - | vedclass

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(C) Let $u = x - 1$. Then $x = u + 1$. The equation becomes: $\sqrt{u + 1 + 3 - 4\sqrt{u}} + \sqrt{u + 1 + 8 - 6\sqrt{u}} = 1$ $\sqrt{u + 4 - 4\sqrt{u

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$x \in [5, 10]$

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(C) Let $u = x - 1$. Then $x = u + 1$. The equation becomes: $\sqrt{u + 1 + 3 - 4\sqrt{u}} + \sqrt{u + 1 + 8 - 6\sqrt{u}} = 1$ $\sqrt{u + 4 - 4\sqrt{u}} + \sqrt{u + 9 - 6\sqrt{u}} = 1$ $\sqrt{(\sqrt{u} - 2)^2} + \sqrt{(\sqrt{u} - 3)^2} = 1$ $|\sqrt{u} - 2| + |\sqrt{u} - 3| = 1$ Let $y = \sqrt{u}$. Since $u \geq 0$,$y \geq 0$. The equation is $|y - 2| + |y - 3| = 1$. This is of the form $|y - a| + |y - b| = |a - b|$,which holds for $y$ between $a$ and $b$ inclusive. Here $a = 2$ and $b = 3$,so $2 \leq y \leq 3$. Substituting $y = \sqrt{u}$,we get $2 \leq \sqrt{u} \leq 3$. Squaring gives $4 \leq u \leq 9$. Since $u = x - 1$,we have $4 \leq x - 1 \leq 9$,which implies $5 \leq x \leq 10$. Thus,$x \in [5, 10]$.

The solution set of the equation $\sqrt{x + 3 - 4\sqrt{x - 1}} + \sqrt{x + 8 - 6\sqrt{x - 1}} = 1$ is

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