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May 26th, 2018, 11:24 AM   #1
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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.
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May 26th, 2018, 11:47 AM   #2
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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 $a - 2$.)
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May 26th, 2018, 12:06 PM   #3
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If $\displaystyle 1\leq a\leq 2$, $\displaystyle \sqrt{a+2\sqrt{a-1}} + \sqrt{a-2\sqrt{a-1}} = \left(1 + \sqrt{a - 1}\right) + \left(1 - \sqrt{a -1}\right) = 2$.
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May 26th, 2018, 12:11 PM   #4
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Maybe I'm just being slow today, but this isn't immediately clear to me:

Quote:
Originally Posted by skipjack View Post
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?
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May 30th, 2018, 04:07 AM   #5
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Thx everyone but visualizing that as 2 is not easy... Solving that square root is not an easy depiction.
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May 30th, 2018, 05:41 AM   #6
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$\displaystyle \sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}$, $1 \le a \le 2$

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 $0 \le t < 1$ ...

$(\sqrt{t}+1) + (1-\sqrt{t}) = 2$
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May 31st, 2018, 12:19 AM   #7
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Thx.
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