Simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
The problem is from a not so well renowned book for calculus in India - Concepts of Functions & Calculus - Vikas Rahi, ISBN 9780070080805. The answer for this question is given as $$\lvert \sqrt{x-1} -1 \rvert + \lvert \sqrt{x-1}+1 \rvert$$ My efforts weren't great at all, I tried rationalizing them but the denominator becomes very similar to the question, which is totally not helpful for simplification.
On
Write $t =\sqrt{x-1}$ then your expression is $$\sqrt{t^2+1+2t} +\sqrt{t^2+1-2t} =|t+1|+|t-1| $$
Notice that $t\geq 0$. So if $t\geq 1$ we get $2t$ and if $t<1$ we get $2$.
On
First, note that \begin{align} (1 \pm \sqrt{x-1})^2 &= 1 \pm 2 \sqrt{x-1} + \sqrt{x-1}^2 \\ &= 1 \pm 2 \sqrt{x-1} + x-1 & x \geq 1 \\ &= x \pm 2 \sqrt{x-1} \end{align}
Thus, since $\sqrt{u^2} = |u|$, \begin{align} f(x) &= \sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}} \\ &= \sqrt{(1-\sqrt{x-1})^2}+\sqrt{(1+\sqrt{x-1})^2} & x \geq 1 \\ &= |1-\sqrt{x-1}|+|1+\sqrt{x-1}| \\ &= \begin{cases} (1-\sqrt{x-1})+(1+\sqrt{x-1}) & 1 \leq x < 2 \\ -(1-\sqrt{x-1})+(1+\sqrt{x-1}) & 2 \leq x \end{cases} \\ &= \begin{cases} 2 & 1 \leq x < 2 \\ 2 \sqrt{x-1} & 2 \leq x \end{cases} \end{align}
On
Just write
\begin{eqnarray*} \sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}} & = & \sqrt{x -1 - 2\sqrt{x -1} +1} +\sqrt{x -1 + 2\sqrt{x -1} +1} \\ & = & \sqrt{(\sqrt{x-1}-1)^2} + \sqrt{(\sqrt{x-1}+1)^2} \\ & = & \begin{cases} 2 & 1 \leq x\leq 2 \\ 2\sqrt{x-1} & x>2 \end{cases} \end{eqnarray*}
On
$$\begin{align}{\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}&=T(x)>0&\end{align}$$
$$T^2(x)=2x+2 \sqrt{x^2-4x+4}$$
$$T^2(x)=2x+2|x-2|$$
$$\begin{align}\color {gold}{\boxed {\color{black}{T(x)=\sqrt{2x+2|x-2|}}}}\end{align}$$
If $x≥2$, then $\begin{align}T(x)&=\sqrt{2x+2x-4}\\ &=2\sqrt{x-1}\end{align}$
If $x<2$, then $\begin{align}T(x)=\sqrt{2x+4-2x}=2.\end{align}$
"Simplify" can easily depend on the beholder's eyes, but one idea (and hint) is:
$$\sqrt{a-b}+\sqrt{a+b}=\frac{2b}{\sqrt{a+b}-\sqrt{a-b}}$$
I don't really think we get a huge simplifaction here but...perhaps something like this is what is meant in that book.