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

How do I simplify this?

  • 4
    What's not simple about that?2011-10-04

2 Answers 2

1

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.

  • 0
    Compare with my [Simple Denesting Formula.](http://math.stackexchange.com/a/90503/242)2011-12-11
2

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.