[Math] How to go about simplifying this complex radical

algebra-precalculusfractions

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:

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

Best Answer

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 $$ \sqrt{ \frac{1 + \frac{ \sqrt{39}}{8}}{2}}. $$