How do you simplify the addition of square root

Question

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

How do I simplify this?

Answer 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.

Answer 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.