My Math Forum Value of an expression
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 May 26th, 2018, 11:24 AM #1 Senior Member     Joined: Jan 2012 Posts: 113 Thanks: 2 Value of an expression Hi, Please let me know: If $\displaystyle 1\leq a\leq 2$ then $\displaystyle \sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}=2$ Is it true? If yes, how? Thx.
 May 26th, 2018, 11:47 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 211 Thanks: 64 Math Focus: Algebraic Number Theory, Arithmetic 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 $\sqrt{(a-2)^2}$ in your working; it's important that $a \leq 2$ so that this equals $2 - a$ rather than $a - 2$.)
 May 26th, 2018, 12:06 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,295 Thanks: 1686 If $\displaystyle 1\leq a\leq 2$, $\displaystyle \sqrt{a+2\sqrt{a-1}} + \sqrt{a-2\sqrt{a-1}} = \left(1 + \sqrt{a - 1}\right) + \left(1 - \sqrt{a -1}\right) = 2$.
May 26th, 2018, 12:11 PM   #4
Senior Member

Joined: Aug 2017
From: United Kingdom

Posts: 211
Thanks: 64

Math Focus: Algebraic Number Theory, Arithmetic Geometry
Maybe I'm just being slow today, but this isn't immediately clear to me:

Quote:
 Originally Posted by skipjack If $\displaystyle 1\leq a\leq 2$, $\displaystyle \sqrt{a+2\sqrt{a-1}} + \sqrt{a-2\sqrt{a-1}} = \left(1 + \sqrt{a - 1}\right) + \left(1 - \sqrt{a -1}\right)$.
Mind explaining how you can see this?

 May 30th, 2018, 04:07 AM #5 Senior Member     Joined: Jan 2012 Posts: 113 Thanks: 2 Thx everyone but visualizing that as 2 is not easy... Solving that square root is not an easy depiction.
 May 30th, 2018, 05:41 AM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 2,761 Thanks: 1416 $\displaystyle \sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}$, $1 \le a \le 2$ let $t = a-1 \implies a = t+1$ $1 \le a \le 2 \implies 0 \le t \le 1$ substitute ... $\sqrt{(t+1) + 2\sqrt{t}} + \sqrt{(t+1) - 2\sqrt{t}}$ $\sqrt{t+2\sqrt{t}+1} + \sqrt{t-2\sqrt{t}+1}$ $\sqrt{(\sqrt{t}+1)^2} + \sqrt{(\sqrt{t}-1)^2}$ $|\sqrt{t} + 1| + |\sqrt{t}-1|$ for $t=1$ ... $|1 + 1| + |1-1| = 2$ for $0 \le t < 1$ ... $(\sqrt{t}+1) + (1-\sqrt{t}) = 2$ Thanks from v8archie
 May 31st, 2018, 12:19 AM #7 Senior Member     Joined: Jan 2012 Posts: 113 Thanks: 2 Thx.

 Tags expression

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post nietzsche Calculus 2 March 31st, 2018 06:37 PM Dacu Number Theory 2 September 20th, 2014 12:59 PM kev085 Algebra 1 April 24th, 2009 04:06 AM woodman5k Algebra 2 October 10th, 2007 04:53 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2018 My Math Forum. All rights reserved.