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$2\sqrt{x} + \sqrt{3}$

How do I simplify this?

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It can’t be simplified any further. – Brian M. Scott Oct 3 '11 at 23:25
Check the edit. I left something out. – David Oct 3 '11 at 23:28
That doesn't make it any better. – joriki Oct 3 '11 at 23:32
Brian's comment still applies, even after your edit. – J. M. Oct 3 '11 at 23:32
What's not simple about that? – mixedmath Oct 4 '11 at 0:47
up vote 1 down vote accepted

Maybe you are searching for something similar to this:

$$\sqrt{a{^+ _-}\sqrt{b}}=\sqrt{\frac{a+c}{2}}{^+ _-}\sqrt{\frac{a-c}{2}}$$


A example:


In your problem:








The final is:


But it needs a better investigation of the existence conditions in these calculus.

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I haven't followed the details of the calculations, but how is the final result supposed to be simpler than $2\sqrt{x} + \sqrt{3}$? – t.b. Nov 11 '11 at 1:36
It's no simpler (I wrote Maybe you are searching for something similar to this:), it's just a transformation and maybe was he is looking for. If he tells not I'll delete if none think this useful. – GarouDan Nov 16 '11 at 13:26
Compare with my Simple Denesting Formula. – Bill Dubuque Dec 11 '11 at 17:49

This seems a simpler derivation to me:

Squaring $a \sqrt x + \sqrt b$ we get $a^2 x + b + 2a \sqrt{bx}$ so $$ a \sqrt x + \sqrt b = \sqrt{a^2 x + b + 2a \sqrt{bx}}. $$

This seems to be a more complicated result to me.

I know, it's all about me.

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