How to write \sqrt{5 + \sqrt{2 sqrt6 }} as \sqrt{a} + \sqrt{b}

[samy]

[FONT=MathJax_Main]Write:

## \sqrt{5+\sqrt{2sqrt6}}##

[/FONT][FONT=MathJax_Main] in terms of \ sqrt{a} +\ sqrt{b}.

[/FONT]my last step is: ##{{5+2^\frac{3} {2} \ 3^\frac{1} {2}}^\frac {1} {2} ##

 

[stapel]

[FONT=MathJax_Main]Write:

## \sqrt{5+\sqrt{2sqrt6}}##

[/FONT][FONT=MathJax_Main] in terms of \ sqrt{a} +\ sqrt{b}.

[/FONT]my last step is: ##{{5+2^\frac{3} {2} \ 3^\frac{1} {2}}^\frac {1} {2} ##

I think you're saying the following:

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The exercise is:

. . . . .\(\displaystyle \mbox{Write }\, \sqrt{\strut 5\, +\, \sqrt{\strut 2\, \sqrt{\strut 6\,}\,}\,}\, \mbox{ in terms of }\, \sqrt{\strut a\,}\, +\, \sqrt{\strut b\,}.\)

I've gotten as far as:

. . . . .\(\displaystyle \left(5\, +\, \dfrac{2^{3/2}}{3^{1/2}}\right)^{1/2}\)

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...but I'm quite unsure about my interpretation of your last line.

When you reply, please include a clear listing of all of your work so far (not just the last line), so we can see what you're doing. Thank you!

 

[samy]

it is: square root {5+2 times square root 6}, thank you.

 

[pka]

it is: square root {5+2 times square root 6}, thank you.

You could have written it that clearly to begin with!

If \(\displaystyle \large\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b}\) then squaring you get
\(\displaystyle \large{5+2\sqrt{6}}=a +2\sqrt{ab}+{b}\) or \(\displaystyle \large \left\{ \begin{array}{l}a+b=5\\ab=6\end{array} \right.\)

Solve for \(\displaystyle a~\&~b\).

POST your results.

 

[samy]

You could have written it that clearly to begin with!

If \(\displaystyle \large\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b}\) then squaring you get
\(\displaystyle \large{5+2\sqrt{6}}=a +2\sqrt{ab}+{b}\) or \(\displaystyle \large \left\{ \begin{array}{l}a+b=5\\ab=6\end{array} \right.\)

Solve for \(\displaystyle a~\&~b\).

POST your results.

Thank you :it is a,b = 2,3

 

[samy]

thank you for your help.

a+b=5 and ab=6 then,solving both together or two numbers, their product is 6 and their sum is 5 then a=2, b=3