I have an exercise that says :
Simplify $\sqrt{16+2\sqrt{55}}$.
Please, I need a vivid explanation. Can anyone help?
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here are some leads to follow https://math.stackexchange.com/questions/1214527/denesting-a-square-root-sqrt7-sqrt14/1454914?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa – David Quinn Apr 08 '18 at 21:41
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$David Quinn$, Thanks for the link. It's really helpful – user9602923 Apr 08 '18 at 22:49
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Possible duplicate of Strategies to denest nested radicals. – Isaac Browne Apr 08 '18 at 23:07
2 Answers
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We have $a^2+2ab+b^2 =(a+b)^2$. If we're lucky, we can use that. Assuming that the $2\sqrt{55}$ term corresponds in some nice way to the $2ab$ term, then most likely we have $a=\sqrt5,b=\sqrt{11}$. And, oh, look: $16=\sqrt5^2+\sqrt{11}^2$.
Arthur
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When I tried it, my final answer was, when a=11,b=5 and when b=11, a=5. And I'm confused which of the values of b am I expected to use. Can I use any of them? – user9602923 Apr 08 '18 at 21:56
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Note that a and b are interchangeable. Whatever is a solution to a is also a solution to b. So, the solutions to a are the solutions to the problem. – Kaynex Apr 08 '18 at 22:16
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@user9602923 Note that the final answer isn't to find $a$ and $b$, it is to find $16+2\sqrt{55}=(\sqrt5+\sqrt{11})^2$. If you instead find $(\sqrt{11}+\sqrt{5})^2$, that's just as valid. – Arthur Apr 08 '18 at 22:25
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Oh, wow. Thanks guys. Kaynex and Arthur. I'm very excited and grateful to you. – user9602923 Apr 08 '18 at 22:46
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$$16+2\sqrt{55} = 11 +5 +2\sqrt{5} \sqrt{11} = ( \sqrt{11}+ \sqrt{5})^2$$
Thus $$\sqrt{16+2\sqrt{55}}=\sqrt{11}+ \sqrt{5}$$
Mohammad Riazi-Kermani
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