How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$

Question

$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$

How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$

Answer 1

$ (\sqrt u + \sqrt v)^2 = u+v + 2\sqrt{u\,v }$

So at first bring in $2$ in front of the radical sign.

$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}$$

Next find two factors for $3$ whose sum is $4$

They are easy to guess. They are $3$ and $1$;

(... or else you need to solve another quadratic equation)

$$ \sqrt{\frac{(\sqrt3 +1)^2}{2}}=\sqrt{ \frac{2(\sqrt3 +1)^2}{4}} $$

where the denominator is made to be $4$ so that while square rooting, $2$ would stay there.

Next manipulate numerator to make them squares of two terms and make denominator also a square

$$\sqrt{ \frac{(\sqrt{2})^2(\sqrt3 +1)^2}{2^2}} $$

that is now ( cancelling out squares and square roots) and multiplying two terms in the numerator:

$$ \dfrac{\sqrt 6 +\sqrt 2}{2}.$$

Answer 2

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$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}=\sqrt{\frac{(1+\sqrt3)^2}{2}}=\frac{1+\sqrt3}{\sqrt2}$$ Can you end it now?

Answer 3

You can use the identity \begin{equation} \sqrt{a+\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} + \sqrt{\frac{a-\sqrt{a^2-b}}{2}}. \end{equation}

Answer 4

Hint: $$\left( \frac{\sqrt{6} + \sqrt{2}}{2}\right)^2 = \frac{6 + 2 \sqrt{6} \sqrt{2} + 2}{4} = ?$$

Answer 5

Assume $$\sqrt{a+\sqrt b}=\sqrt c+\sqrt d$$ where $a,b,c,d$ are all rational.

Then

$$a+\sqrt b=c+2\sqrt{cd}+d.$$

By identification

$$\begin{cases}a=c+d,\\b=4cd.\end{cases}$$

From this,

$$4c^2+4cd-4ac=4c^2-4ac+b=0.$$

This is a quadratic equation in $c$, with roots

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

So for a rational solution to exist, $a^2-b$ must be a perfect rational square.

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With $a=2,b=3$,

$$\sqrt{2+\sqrt3}=\sqrt{\frac12}+\sqrt{\frac32}=\frac{\sqrt2+\sqrt6}2.$$

Answer 6

If you want to change $\sqrt{2+ \sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$, then go through the following calculation steps from bottom to top. If you want to change $\dfrac{\sqrt{6}+\sqrt{2}}{2}$ into $\sqrt{2+ \sqrt{3}}$, then go through the following calculation steps from top to bottom. Either way we get equality $\sqrt{2+ \sqrt{3}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}$.

\begin{align*} \dfrac{\sqrt{6}+\sqrt{2}}{2} &= \sqrt{\left(\dfrac{\sqrt{6}+\sqrt{2}}{2}\right)^2} && a = \sqrt{a^2} \text{ for } a \geq 0 \\&= \sqrt{\dfrac{(\sqrt{6}+\sqrt{2})^2}{2^2}} && \left(\dfrac{a}{b}\right)^2 = \dfrac{a^2}{b^2} \\&= \sqrt{\dfrac{(\sqrt{6})^2+2\cdot\sqrt{2}\cdot\sqrt{6}+(\sqrt{2})^2}{4}} && (a+b)^2 = a^2+2\cdot a\cdot b+b^2 \\&= \sqrt{\dfrac{6+2\cdot\sqrt{2}\cdot\sqrt{6}+2}{4}} && a = (\sqrt{a}) \text{ for } a \geq 0 \\&= \sqrt{\dfrac{6+\sqrt{4}\cdot\sqrt{2}\cdot\sqrt{6}+2}{4}} && 2 = \sqrt{4} \\&= \sqrt{\dfrac{6+\sqrt{4\cdot 2\cdot 6}+2}{4}} && \sqrt{a}\cdot \sqrt{b}\cdot \sqrt{c} = \sqrt{a\cdot b\cdot c} \\&= \sqrt{\dfrac{6+\sqrt{48}+2}{4}} && 4\cdot 2\cdot 6=8\cdot6=48 \\&= \sqrt{\dfrac{6+\sqrt{3\cdot 16}+2}{4}} && 48 = 3 \cdot 16 \\&= \sqrt{\dfrac{6+\sqrt{3}\cdot \sqrt{16}+2}{4}} && \sqrt{a\cdot b} = \sqrt{a}\cdot \sqrt{b} \\&= \sqrt{\dfrac{6+\sqrt{3}\cdot 4+2}{4}} && \sqrt{16} = \sqrt{4^2} = 4 \\&= \sqrt{\dfrac{6+4 \cdot \sqrt{3}+2}{4}} && a \cdot b = b\cdot a \\&= \sqrt{\dfrac{8+4 \cdot \sqrt{3}}{4}} && 6+2=8 \\&= \sqrt{\dfrac{8}{4}+\dfrac{4 \cdot \sqrt{3}}{4}} && \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c} \\&= \sqrt{2+ \sqrt{3}} \end{align*}