Evaluate sqrt{x + 2sqrt{x - 1}} + sqrt{x - 2s | vedclass

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(A) Let $y = x - 1$. Then $x = y + 1$. Since $x \ge 1$,$y \ge 0$.Substituting this into the expression: $\sqrt{y + 1 + 2\sqrt{y}} + \sqrt{y + 1 - 2\sq

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$2$,if $1 \le x \le 2$

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(A) Let $y = x - 1$. Then $x = y + 1$. Since $x \ge 1$,$y \ge 0$.Substituting this into the expression: $\sqrt{y + 1 + 2\sqrt{y}} + \sqrt{y + 1 - 2\sqrt{y}}$.This simplifies to $\sqrt{(\sqrt{y} + 1)^2} + \sqrt{(\sqrt{y} - 1)^2}$.$= |\sqrt{y} + 1| + |\sqrt{y} - 1|$.Since $\sqrt{y} + 1$ is always positive,this is $\sqrt{y} + 1 + |\sqrt{y} - 1|$.Case $1$: If $0 \le y \le 1$ (i.e.,$1 \le x \le 2$),then $|\sqrt{y} - 1| = 1 - \sqrt{y}$.Expression $= \sqrt{y} + 1 + 1 - \sqrt{y} = 2$.Case $2$: If $y > 1$ (i.e.,$x > 2$),then $|\sqrt{y} - 1| = \sqrt{y} - 1$.Expression $= \sqrt{y} + 1 + \sqrt{y} - 1 = 2\sqrt{y} = 2\sqrt{x - 1}$.

Evaluate $\sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}}$ for $x \ge 1$.

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