Dear Uncle Colin

How does $\sqrt{9 - \sqrt{17}} = \frac{\sqrt{34}-\sqrt{2}}{2}$? I tried applying a formula, but I couldn’t make it work.

- Roots Are Dangerous, It’s Chaotic A-Level Simplification

Hi, RADICALS, and thank you for your message!

Square roots of square roots are not usually trivial, but this one can (clearly) be simplified.

Here’s how I’d do it

Let’s suppose that our expression can be written as the sum of simple square roots:

$\sqrt{9-\sqrt{17}} = \sqrt{a} - \sqrt{b}$ - or, better:

$9 - \sqrt{17} = \br{\sqrt{a} - \sqrt{b}}^2 = a + b - 2\sqrt{ab}$

Now, the integer and root parts of that need to match up:

  • $9 = a + b$
  • $\sqrt{17} = 2\sqrt{ab}$, or $17 = 4ab$.

Those aren’t nice to solve simultaneously, but if $a = 9-b$, we can substitute into the second equation to get $17 = 4(9-b)b$, or $4b^2 - 36b + 17 = 0$.

That factorises: $(2b - 1)(2b - 17) = 0$, so $b = \frac{1}{2}$ or $b = \frac{17}{2}$.

Corresponding to those, $a = \frac{17}{2}$ or $\frac{1}{2}$ - so (according to the sum), $a = \frac{17}{2}$ and $b = \frac{1}{2}$, in either order. In fact, we know $\sqrt{a}-\sqrt{b} > 0$, so the order I’ve given there is the right one.

Now to simplify

We’ve got $\sqrt{9-\sqrt{17}} = \sqrt{\frac{17}{2}} - \sqrt{\frac{1}{2}}$.

Combining that right-hand fraction gives $\frac{\sqrt{17} - 1}{\sqrt{2}}$.

Rationalising, by multiplying through by $\sqrt{2}$ on top and bottom, gives $\frac{\sqrt{34}-\sqrt{2}}{2}$, as required.

Hope that helps!

- Uncle Colin