# How do I go about simplifying this complex radical?

I'm having a difficult time trying to simplify the radical below. When I type the radical to the left into my calculator, I arrive at the correct answer to the right. But I cannot seem to figure out how to simplify the radical on the left by hand. I've googled the heck out of it, but perhaps I'm using an incorrect search phrase, because I cannot find a similar example. Can someone please either point in the right direction or give me a step by step system for simplifying the radical on the left to arrive to the solution on the right?

radical image:

$$\sqrt{\frac{1+\frac{\sqrt{39}}{8}}{2}}=\frac{\sqrt{26}+\sqrt 6}{8}$$

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Do you seek to verify the equality, or instead to derive the denested closed form? Your question asks the latter, but the answer you accepted seems to imply the former? –  Bill Dubuque Oct 25 '12 at 18:14

## 3 Answers

How can we make these two expressions look a little more alike?

Well, the right hand side is not inside a square root; let's change that using the fact that $\sqrt{a^2}=a$ when $a>0$. So we can write the right hand side as $$\sqrt{\frac{(\sqrt{26}+\sqrt{6})^2}{8^2}}.$$ Now we multiply out. $$\sqrt{ \frac{32 + 2\sqrt{6}\sqrt{26} }{ 64} }$$ The most complicated part here is the $\sqrt{6}\sqrt{26}$ and this is not on the left hand side. So how can we rewrite $\sqrt{6}\sqrt{ 26}$? Well this is $\sqrt{2}\sqrt{3}\sqrt{2}\sqrt{13}=2\sqrt{3}\sqrt{13}=2\sqrt{39}$. Filling this in, we now have $$\sqrt{ \frac{32 + 4\sqrt{39} }{ 64} }.$$ If we divide above and below by 32, we now get $$\sqrt{ \frac{1 + \frac{ \sqrt{39}}{8}}{2}}.$$

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That's the wrong direction. Given the LHS, the OP seeks to discover the "simpler" denested RHS form. But you already assume as given the denested answer. –  Bill Dubuque Oct 25 '12 at 14:52

Method One: This is another technique for simplifying nested radicals of this form if they can be simplified. It is based on the following binomial formula $a^2+2ab+b^2$. First let's put the nested radical in this form $\sqrt{\frac{1}{2}+\frac{\sqrt{39}}{16}}$. Set $$a^2+b^2=\frac{1}{2}$$ and $$2ab=\frac{\sqrt{39}}{16}$$ corresponding to the each real number underneath the radical sign. The second equation results in $b=\frac{\sqrt{39}}{32a}$. Substitute b into the first equation to arrive at the following equation $a^2+(\frac{\sqrt{39}}{32a})^2=\frac{1}{2}$; $a^2+\frac{39}{1,024a^2}=\frac{1}{2}$. Multiply all terms by $a^2$ and set the equation equal to zero. So, $$a^4-\frac{1}{2}a^2+\frac{39}{1,024}=0$$ Use the quadratic formula to arrive at $a_1^2=\frac{26}{64}$ and $a_2^2=\frac{6}{64}$ or $a_1=\pm\frac{\sqrt26}{8}$ and $a_2=\pm\frac{\sqrt6}{8}$. If we add the two positive roots $a_1$ and $a_2$ and we can see that $$\sqrt{\frac{1}{2}+\frac{\sqrt{39}}{16}}=\frac{\sqrt{26}+\sqrt 6}{8}$$ The equality $\sqrt{\frac{1}{2}-\frac{\sqrt{39}}{16}}$=$\frac{\sqrt{26}-\sqrt 6}{8}$ is computed the same way. If the result of these computations do not denest the nested radical nicely the result will be even for nested than the original nested radical. Nevertheless, the result will still be equal to the original nested radical.

Method Two: Another similar method where we use the quadratic equation but only to degree two not four is as follows. Set the nested radical as the sum of two square roots so that $$\sqrt{\frac{1}{2}+\frac{\sqrt{39}}{16}}=(\sqrt{a}+\sqrt{b})$$ Then square both sides so that $$\frac{1}{2}+\frac{\sqrt{39}}{16}=a+2\sqrt{a}\sqrt{b}+b$$ Set (1) $$a+b=\frac{1}{2}$$ and set (2) $$2\sqrt{a}\sqrt{b}=\frac{\sqrt{39}}{16}$$ Square both sides of (2) to get $4ab= \frac{{39}}{256}$ and solve for $b$ to get $b=\frac{39}{1,024a}$. Putting $b$ into (1) gives $a+\frac{39}{1,024a}=\frac{1}{2}$. Multiply all terms by $a$ and convert to the quadratic equation $$a^2-\frac{1}{2}a+\frac{39}{1,024}=0$$ Solving the quadratic gives $a=\frac{26}{64}$or $a=\frac{6}{64}$. As a result at the same time $b=\frac{26}{64}$or $b=\frac{6}{64}$. Replacing $a$ and $b$ in the sum or square roots formula above gives $\frac{\sqrt{26}+\sqrt 6}{8}$.

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Scaling both sides by $\:\sqrt{16} = 4\:$ we wish to derive the following radical denesting

$$\sqrt{8 + \sqrt{39}}\ =\ \frac{\sqrt{6}+\sqrt{26}}2$$

Given only the LHS  nested  radical, without any hint of the form of the RHS  denested  answer,  we can easily compute the denesting using a very simple rule that I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace}$

Recall $\rm\: w = a + b\sqrt{n}\:$ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2$

and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

Here $\:8+\sqrt{39}\:$ has norm $= 5^2.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 5\$ yields $\ 3+\sqrt{39}\:$

and this has $\rm\ \sqrt{trace}\: =\: \sqrt{6},\ \ so,\ \ \ \color{brown}{dividing\ it\ out}\$ of this yields

$$\frac{3+\sqrt{39}}{\sqrt{6}}\, =\, \frac{3\sqrt{6}+3\sqrt{26}}{6}\,=\,\frac{\sqrt{6}+\sqrt{26}}2$$

For many further examples see my prior posts on denesting.

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