Lately, I have been doing a lot of math involving square roots of numbers added to square roots, in the form of \(\sqrt{A+B\sqrt{C}}\), this is called a “nested radical.” Normally, you would not be able to simplify any further, unless there was a common factor of both that could be removed, or if both items had the same number under the radical (eg. \(\sqrt{4\sqrt7+12\sqrt7}=\sqrt{16\sqrt7}=4\sqrt[4]{7}\) ).

I can take a number like \((1+\sqrt5)^2\) and work out it is equal to \(6 + 2\sqrt5\). Unfortunately, it is very difficult to recognize that \(1+\sqrt5\) is the square root (or even a factor) of that. Even harder, trying to find an equation such as \(\sqrt{469+144\sqrt5}\).

Recently, I found a method of breaking down the equation into its square root equation.

With an equation in the form of \(\sqrt{A+B\sqrt{C}}\), where C contains no perfect square factors, taking the formula \(D^2 = A^2 – (B^2)*C\), if D is a rational number, you will be able to factor the equation, otherwise it is simplified as much as is possible.

Then you will be able to get the formula: \(\sqrt{A+B\sqrt{C}} = \pm \left(\sqrt{\frac{A-D}{2}} + sign(A*B)*\sqrt{\frac{A+D}{2}}\right)\), where sign(A*B) is the positive or negative value of A*B, it will always give you 0, +1, or -1.

So to find \(\sqrt{469+144\sqrt5}\), we find \(469^2 – 144^2 * 5 = 116281\). Since 116281 = 341*341, we can plug D=341 into the formula to get:

$$\begin{align} \sqrt{\frac{469-341}{2}} + 1*\sqrt{\frac{469+341}{2}}&=\sqrt{\frac{128}{2}} + \sqrt{\frac{810}{2}}\\

&=\sqrt{64} + \sqrt{405} \\

&= 8 + \sqrt{81*5} \\

&= 8+9\sqrt{5}\end{align}$$

So, \(\sqrt{469+144\sqrt5} = (8+9\sqrt{5}) \text{ and } (-8-9\sqrt{5})\).

To find \(\sqrt{53-12\sqrt{11}}\), we find \(53^2 – 12^2 * 11 = 1225\). Since 1225 = 35*35, we can plug D=35 into the formula to get:

$$\begin{align*} \sqrt{\frac{53-35}{2}} + -1*\sqrt{\frac{53+35}{2}} &=\sqrt{\frac{18}{2}} – \sqrt{\frac{88}{2}}\\

&=\sqrt{9} – \sqrt{44} \\

&= 3 – \sqrt{4*11} \\

&= 3-2\sqrt{11} \end{align*}$$

So, \(\sqrt{53-12\sqrt11} = (3-2\sqrt{11}) \text{ and } (-3+2\sqrt{11})\).

If A and B have the same sign, the answer will have both terms the same sign ({+,+} and {-,-} are equally valid). If A and B have different signs, the terms in the answer will have different signs (± and ∓). Often you will need to determine the complete answer that equates to positive, after all, it is hard to have a real object with a negative length.

An interesting correlation: \((X-Y)^2 = (Y-X)^2\), but remember \(\sqrt{X-Y} \not= \sqrt{Y-X}\). Double check your signs by re-squaring your answer.

Note: If A or B is a zero, then this technique doesn’t really work well. Example:

$$\begin{align*} \sqrt{4+0\sqrt6} &=\sqrt{\frac{4-2}{2}} + \sqrt{\frac{4+2}{2}}\\

&=\sqrt{\frac{2}{2}} + \sqrt{\frac{6}{2}}\\

&=\sqrt1 + \sqrt3 \\

&= 1 + \sqrt3 \\

\end{align*}$$

Which is obviously not correct. \((1+\sqrt3)^2=4+2\sqrt3\)