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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.

Thanks in advance!

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    $\begingroup$ Apply $a^2-b^2=(a+b)(a-b)$, $\endgroup$ – choco_addicted Mar 20 '18 at 8:26
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Use the fact that $\sqrt{a+b}*\sqrt{a-b}=\sqrt{(a-b)(a+b)}$, and then use @choco_addicted's comment.

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  • $\begingroup$ 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 $$8+2|a|$$ $\endgroup$ – klamsi Mar 20 '18 at 8:59
  • $\begingroup$ 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 $\sqrt{a^2}=|a|$. $\endgroup$ – Mrtny Mar 20 '18 at 9:19
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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 $$8+2|a|.$$

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

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