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

How do I simplify this?

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    $\begingroup$ It can’t be simplified any further. $\endgroup$ Oct 3 '11 at 23:25
  • $\begingroup$ Check the edit. I left something out. $\endgroup$
    – David
    Oct 3 '11 at 23:28
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    $\begingroup$ That doesn't make it any better. $\endgroup$
    – joriki
    Oct 3 '11 at 23:32
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    $\begingroup$ Brian's comment still applies, even after your edit. $\endgroup$ Oct 3 '11 at 23:32
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    $\begingroup$ What's not simple about that? $\endgroup$
    – davidlowryduda
    Oct 4 '11 at 0:47
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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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Maybe you are searching for something similar to this:

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

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

A example:

$\sqrt{5{^+_-}\sqrt{24}}=\sqrt{\frac{5+1}{2}}{^+_-}\sqrt{\frac{5-1}{2}}=\sqrt{3}{^+_-}\sqrt{2}$

In your problem:

$2\sqrt{x}+\sqrt{3}=\sqrt{3}+\sqrt{4x}$

$a+c=2\times3$

$a-c=2\times4x$

So,

$a=4x+3$

$c=3-4x$

$b=a^2-c^2=(a-c)(a+c)=8x\times6=48x=3x\times4^2$

The final is:

$$2\sqrt{x}+\sqrt{3}=\sqrt{4x+3+4\sqrt{3x}}$$

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

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    $\begingroup$ I haven't followed the details of the calculations, but how is the final result supposed to be simpler than $2\sqrt{x} + \sqrt{3}$? $\endgroup$
    – t.b.
    Nov 11 '11 at 1:36
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    $\begingroup$ 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. $\endgroup$
    – GarouDan
    Nov 16 '11 at 13:26
  • $\begingroup$ Compare with my Simple Denesting Formula. $\endgroup$ Dec 11 '11 at 17:49

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