# Ask Uncle Colin: A Radical Feast

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