<s>[INST] Problem in Rationalization The Question is :
If $1 < a <2$ then 
$\sqrt{a - 2 \sqrt{a - 1}} - \sqrt{a + 2 \sqrt{a - 1}}$ can be:
A) $2$
B)$-2 \sqrt{a-1}$
C) $0$ 
D) $\sqrt{a-1}$
I have tried solving it a number of times but failed every time. 
I took $\sqrt{a-1}$ as b then tried to rationalise by multiplying with unity ($\sqrt{a-2b}$ with $\sqrt{a+2b}$ and vice versa). However, after a few steps I always arrived at the question itself not the answer. 
Can someone please help me with this sum?
 [/INST] $$k=\sqrt{a-2\sqrt{a-1}}-\sqrt{a+2\sqrt{a-1}}$$
square both sides and get:
$$k^2=2a-2\sqrt{a^2-4(a-1)}=2a-2|a-2|$$
once $1<a<2$ then $|a-2|=2-a$, so
$$k^2=2a-2(2-a)=4a-4\to k=\pm2\sqrt{a-1}$$
but $k<0$ because $\sqrt{a-2\sqrt{a-1}}<\sqrt{a+2\sqrt{a-1}}$, so
$$k=-2\sqrt{a-1}$$
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