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May 26th, 2018, 12:24 PM  #1 
Senior Member Joined: Jan 2012 Posts: 123 Thanks: 2  Value of an expression
Hi, Please let me know: If $\displaystyle 1\leq a\leq 2$ then $\displaystyle \sqrt{a+2\sqrt{a1}}+\sqrt{a2\sqrt{a1}}=2$ Is it true? If yes, how? Thx. 
May 26th, 2018, 12:47 PM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 287 Thanks: 88 Math Focus: Number Theory, Algebraic Geometry 
Yes, it's true. Just try squaring the expression on the left hand side and see what happens. (At one point you'll probably end up with a term like $\sqrt{(a2)^2}$ in your working; it's important that $a \leq 2$ so that this equals $2  a$ rather than $a  2$.)

May 26th, 2018, 01:06 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,098 Thanks: 1905 
If $\displaystyle 1\leq a\leq 2$, $\displaystyle \sqrt{a+2\sqrt{a1}} + \sqrt{a2\sqrt{a1}} = \left(1 + \sqrt{a  1}\right) + \left(1  \sqrt{a 1}\right) = 2$.

May 26th, 2018, 01:11 PM  #4 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 287 Thanks: 88 Math Focus: Number Theory, Algebraic Geometry 
Maybe I'm just being slow today, but this isn't immediately clear to me: Mind explaining how you can see this? 
May 30th, 2018, 05:07 AM  #5 
Senior Member Joined: Jan 2012 Posts: 123 Thanks: 2 
Thx everyone but visualizing that as 2 is not easy... Solving that square root is not an easy depiction.

May 30th, 2018, 06:41 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449 
$\displaystyle \sqrt{a+2\sqrt{a1}}+\sqrt{a2\sqrt{a1}}$, $1 \le a \le 2$ let $t = a1 \implies a = t+1$ $1 \le a \le 2 \implies 0 \le t \le 1$ substitute ... $\sqrt{(t+1) + 2\sqrt{t}} + \sqrt{(t+1)  2\sqrt{t}}$ $\sqrt{t+2\sqrt{t}+1} + \sqrt{t2\sqrt{t}+1}$ $\sqrt{(\sqrt{t}+1)^2} + \sqrt{(\sqrt{t}1)^2}$ $\sqrt{t} + 1 + \sqrt{t}1$ for $t=1$ ... $1 + 1 + 11 = 2$ for $0 \le t < 1$ ... $(\sqrt{t}+1) + (1\sqrt{t}) = 2$ 
May 31st, 2018, 01:19 AM  #7 
Senior Member Joined: Jan 2012 Posts: 123 Thanks: 2 
Thx.


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