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I'm trying to prove that

$$ \sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}} $$ With $$ C=\sqrt{A^2 - B} $$

How can I handle this?

Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.

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    $\begingroup$ Did you try squaring both sides? $\endgroup$
    – kingW3
    Nov 23, 2015 at 16:24
  • $\begingroup$ Yep, I know that it works, but I want to derive the result without using the r.h.s. $\endgroup$
    – james42
    Nov 23, 2015 at 16:26
  • $\begingroup$ So, you aren't really looking for a proof, you are looking for a derivation? $\endgroup$ Nov 23, 2015 at 16:28
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    $\begingroup$ Usually, the intuition behind a formula like this will come from some exterior reason (such as a geometric reason). For example, you could try seeing this in terms of a rectangle whose diagonal is $\sqrt{A+\sqrt{B}}$ and compute the length of the diagonal in two different ways. $\endgroup$ Nov 23, 2015 at 16:31
  • $\begingroup$ @ThomasAndrews Exactly! :) $\endgroup$
    – james42
    Nov 23, 2015 at 16:32

7 Answers 7

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Well assume that $\sqrt{a+\sqrt{b}}$ can be written as sum of 2 square roots $$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}\\a+\sqrt{b}=x+y+\sqrt{4xy}\\a=x+y\\b=4xy\\x=a-y\\b=4(a-y)y\\b=4ay-4y^2\\4y^2-4ay+b=0\\y_{1,2}=\frac{4a\pm\sqrt{16a^2-16b}}{8}\\y_{1,2}=\frac{a\pm\sqrt{a^2-b}}{2}\\x_{1,2}=\frac{a\mp\sqrt{a^2-b}}{2}$$ Now it $x_1=y_2$ and $x_2=y_1$ so that doesn't matter at all,now set $C=\sqrt{a^2-b}$ and you get your formula

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  • $\begingroup$ this is the way the formula is obtained. Other answers are mere verifications. +1 $\endgroup$
    – Paramanand Singh
    Nov 24, 2015 at 4:41
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Before starting, let us note that the formula is not really $a$ nested radicals formula, because it suffices to replace $\sqrt{B}$ by $B,$ and replace $C^{2}=A^{2}-B,$ by $C^{2}=A^{2}-B^{2}.$ So let me show you how starting from the l.h.s. written as \begin{equation*} \sqrt{A+B} \end{equation*} to arrive to the r.h.s written as \begin{equation*} \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}},\ \ \ \ with\ \ \ C=\sqrt{% A^{2}-B^{2}}. \end{equation*}

Consider a rectangle triangle with sides $A$, $B$, and $C$ such that (pythagor theorem) \begin{equation*} A^{2}=B^{2}+C^{2}\ \ \ \ \ \ (\ast ) \end{equation*} then \begin{eqnarray*} B^{2} &=&A^{2}-C^{2} \\ B^{2} &=&(A+C)(A-C) \\ B^{2} &=&4\frac{(A+C)}{2}\frac{(A-C)}{2} \end{eqnarray*} then \begin{equation*} B=2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \end{equation*} and by adding $A$ to both sides \begin{equation*} A+B=A+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \end{equation*} However, note that \begin{equation*} A=\frac{A+C}{2}+\frac{A-C}{2} \end{equation*} then \begin{eqnarray*} A+B &=&A+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \\ &=&\frac{A+C}{2}+\frac{A-C}{2}+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \\ &=&\left( \sqrt{\frac{A+C}{2}}\right) ^{2}+\left( \sqrt{\frac{A-C}{2}}% \right) ^{2}+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \\ A+B &=&\left( \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right) ^{2} \end{eqnarray*} and then \begin{equation*} \sqrt{A+B}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}} \end{equation*} where $C$ is given by (*), that is, $C=\sqrt{A^{2}-B^{2}}.$

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  • $\begingroup$ @Thomas you are right, it is a typo thanks $\endgroup$ Nov 24, 2015 at 4:29
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In what follows, I will show how to start from the l.h.s. written as \begin{equation*} \sqrt{A+\sqrt{B}} \end{equation*} and arrive to the r.h.s written as \begin{equation*} \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}},\ \ \ \ with\ \ \ C=\sqrt{A^{2}-B}. \end{equation*}

To this end, make use of the following standard identities of algebra \begin{eqnarray*} x^{2}-y^{2} &=&(x+y)(x-y) \\ (x+y)^{2} &=&x^{2}+y^{2}+2xy \end{eqnarray*} Note that this second identity can be written as follows \begin{equation} xy=\frac{1}{2}(x+y)^{2}-\frac{1}{2}x^{2}-\frac{1}{2}y^{2} \tag{K} \end{equation} Consider a rectangle triangle with the longest side $A$ and the others are $ \sqrt{B,}$ and $C.$ So, by the Pythagorean theorem one has,$\ $ \begin{equation} A^{2}=B+C^{2}, \tag{P} \end{equation} then$\ \sqrt{B}=\sqrt{A^{2}-C^{2}},$ therefore, \begin{eqnarray*} \sqrt{A+\sqrt{B}} &=&\sqrt{A+\sqrt{A^{2}-C^{2}}} \\ &=&\sqrt{A+\sqrt{(A+C)(A-C)}} \\ &=&\sqrt{A+\sqrt{A+C}\sqrt{A-C}},\ let\ x=\sqrt{A+C},\ and\ y=\sqrt{A-C},\ and\ apply\ (K) \\ &=&\sqrt{A+\frac{1}{2}\left( \sqrt{A+C}+\sqrt{A-C}\right) ^{2}-\frac{1}{2}% \left( \sqrt{A+C}\right) ^{2}-\frac{1}{2}\left( \sqrt{A-C}\right) ^{2}} \\ &=&\sqrt{A+\frac{1}{2}\left( \sqrt{2}\sqrt{\frac{A+C}{2}}+\sqrt{2}\sqrt{% \frac{A-C}{2}}\right) ^{2}-\frac{1}{2}(A+C)-\frac{1}{2}\left( A-C\right) } \\ &=&\sqrt{A+\frac{1}{2}(\sqrt{2})^{2}\left( \sqrt{\frac{A+C}{2}}+\sqrt{\frac{% A-C}{2}}\right) ^{2}-A} \\ &=&\sqrt{\left( \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right) ^{2}} \\ \sqrt{A+\sqrt{B}} &=&\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}},\ \ \ with\ C=% \sqrt{A^{2}-B},\ from\ (P). \end{eqnarray*}

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$$A+\sqrt{B}=\frac{A+C}{2}+\frac{A-C}{2}+\sqrt{A^2-(A^2-B)}$$ $$={\sqrt{\frac{A+C}{2}}}^2+2\sqrt{\frac{(A+\sqrt{A^2-B})(A-\sqrt{A^2-B})}{4}}+{\sqrt{\frac{A-C}{2}}}^2$$ $$={\sqrt{\frac{A+C}{2}}}^2+2\sqrt{\frac{(A+C)(A-C)}{4}}+{\sqrt{\frac{A-C}{2}}}^2$$ $$=\left(\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right)^2$$

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$$A+\sqrt{B}=\frac{A}{2}+\frac{C}{2}+\frac{A}{2}-\frac{C}{2}+\sqrt{B},$$ where $C=\sqrt{A^2-B}$. But $$\sqrt{B}=\sqrt{A^2-(A^2-B)}=\frac{2}{2}\sqrt{(A+\sqrt{A^2-B})(A-\sqrt{A^2-B})}=2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}.$$ Then $$A+\sqrt{B}=\frac{A+C}{2}+\frac{A-C}{2}+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}=\left(\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right)^2.$$ Taking square roots gives the answer.

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Here's an awesome proof I discovered right now :

Consider the quadratic :

$x^2-Ax+\frac{B}{4}=0$ with the roots $x_1$ and $x_2$ .

I have chosen this because of the nice determinant $A^2-4 \cdot \frac{B}{4}=A^2-B=C^2$

Now let's look at the roots :

$$x_{1,2}=\frac{A \pm C}{2}$$ wow exactly those numbers.

Now use Viete's relations to get : $x_1+x_2=A$ and $x_1\cdot x_2=\frac{B}{4}$

The problem is now equivalent with :

$$\sqrt{x_1}+\sqrt{x_2}=\sqrt{x_1+x_2+2\sqrt{x_1x_2}}$$ which should be obvious .

This is very simple but I felt a great joy when I found it . This is the beauty of mathematics :D.

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I can derive the left side from the right, not the other way around.

The right side is the sum of the two positive roots (assuming $A^2>B$) of:

$$x^4-Ax^2+\frac{B}{4}=0$$

But this polynomial factors as:

$$x^4-Ax^2+\frac{B}{4}=\left(x^2+\frac{\sqrt{B}}2\right)^2-(A+\sqrt{B})x^2\\ = \left(x^2-\sqrt{A+\sqrt{B}}x+\frac{\sqrt{B}}2\right) \left(x^2+\sqrt{A+\sqrt{B}}x+\frac{\sqrt{B}}2\right)$$

Note that the positive values cannot be roots to the right factor, so they have to be roots of the left factor, and therefore their sum is $\sqrt{A+\sqrt{B}}$.


Reversing the direction is uglier:

$$A+\sqrt{B}$$ is the larger root of:

$$x^2-2Ax + C^2 = 0$$

Therefore $w=\sqrt{A+\sqrt{B}}$ is the largest root of $$x^4-2Ax^2+C^2 = 0$$

Now $$x^4-2Ax^2+ C^2= \left(x^2+C\right)^2-2(A+C)x^2\\ =\left(x^2-\sqrt{2(A+C)}x+C\right)\left(x^2+\sqrt{2(A+C)}x+C\right)$$

It can't be a root of the right side, which is positive at $w$, so we have: $$w^2-\sqrt{2(A+C)}w+C=0$$

We can also factor the above by completing the square the other way:

$$x^4-2Ax^2+C^2 = (x^2-C)^2 - 2(A-C)x^2 =\\ =\left(x^2-\sqrt{2(A-C)}x-C\right)\left(x^2+\sqrt{2(A-C)}x-C\right)$$

The positive roots of this polynomial have to be split on the two sides, and we can see since that $\sqrt{A+\sqrt{B}}$ must be the largest root, it must be a root of the left factor.

A common root $u=\sqrt{A+\sqrt{B}}$ of the two equations: $$w^2 - \sqrt{2(A+C)}w + C=0\\ w^2-\sqrt{2(A-C)}w - C=0$$

Thus, adding, then dividing by $w$ and re-arranging, we get: $$2u = \sqrt{2(A+C)} + \sqrt{2(A-C)}$$

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