# Simplifying nested radicals? $\left(\sqrt{4+\sqrt{16-a^2}}+\sqrt{4-\sqrt{16-a^2}}\right)^2$

I got the following term I'd like to simplify $$\left(\sqrt{4+\sqrt{16-a^2}}+\sqrt{4-\sqrt{16-a^2}}\right)^2.$$

My Approach was to use the binomial formula. Therefore I'm currently at (hope thats correct): $$8+2×\left(\sqrt{4+\sqrt{16-a^2}}\right)×\left(\sqrt{4-\sqrt{16-a^2}}\right).$$

But now I'm stuck. Any hints or suggestions how to proceed? I had a look at denesting radicals but that doesn't help me yet.

• Apply $a^2-b^2=(a+b)(a-b)$, – choco_addicted Mar 20 '18 at 8:26

Use the fact that $\sqrt{a+b}*\sqrt{a-b}=\sqrt{(a-b)(a+b)}$, and then use @choco_addicted's comment.

• So: $$8+2×\left(\sqrt{(4+\sqrt{16-a^2})×(4-\sqrt{16-a^2}})\right).$$ $$8+2*\sqrt{16-16+a^2}$$ Hence: $$8+2a$$ Seems like I missed something... as It should be $$8+2|a|$$ – klamsi Mar 20 '18 at 8:59
• This is because the square root is positive by convention. Thus, if $a$ was negative, the answer $\sqrt{a^2}=a$ would be negative, going against the convention. This means $\sqrt{a^2}=|a|$. – Mrtny Mar 20 '18 at 9:19

This can be done mentally.

Expanding the square of the sum, the individual squares indeed yield $$8$$, and the double product is the square root of a difference of squares $$\sqrt{4^2-(16-a^2)}$$.

The answer is $$8+2|a|.$$

• So $8\pm2a$ would be your final answer? – klamsi Mar 20 '18 at 8:32
• @klamsi: mh, no, actually $8+2|a|$. – Yves Daoust Mar 20 '18 at 8:33