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## Main Question or Discussion Point

How would you approach something like this?

[tex]\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}[/tex]

[tex]\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}[/tex]

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- #1

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How would you approach something like this?

[tex]\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}[/tex]

[tex]\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}[/tex]

- #2

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Try turning the radicals into exponents.

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- #3

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Let [tex]17-4\sqrt{15} = x^2[/tex]How would you approach something like this?

[tex]\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}[/tex]

There are two possibilities for the form of x:

[tex]x = A + B\sqrt{15}[/tex]

[tex]x = A\sqrt{3} + B\sqrt{5}[/tex]

The first gives 2AB = -4, A^2 + 15 B^2 = 17

I don't see any solutions to that. Try the other form ;-)

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- #5

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If you go down that path you'll end up with a 32nd order polynomial. It almost certainly will not have integer roots, so I've no idea how you would go about finding factors.

(Update: as it happens it does have an integer root. The puzzle setter was unnecessarily kind.)

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- #6

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Would you mind explaining why you choose the deepest radical?If you go down that path you'll end up with a 32nd order polynomial. It almost certainly will not have integer roots, so I've no idea how you would go about finding factors.

- #7

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I assumed that whoever set this puzzle knows there's a neat answer. The only chance I could see of finding a neat answer is to start with the innermost part and simplify that. Presumably that blends nicely with the next enclosing part, allowing the process to be repeated.Would you mind explaining why you choose the deepest radical?

The answer, incidentally, is an integer.

- #8

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It was letting a+√b = √(4+2√(3)), squaring both sides and doing the algebra gave a much simpler expression of 1+√(3)

I used that method and tried experimenting with different variations of it (using a negative sign and adding different coefficients in front of √(b) - I'm not sure how they figured out that formula/method and am very interested in how these formulas are found) but my answer is always off by 2. My calculator says that the expression (the deepest one) should simplify to 2√(3) - √(5)

By the way, bit an elementary question, when we're squaring the equation, aren't we changing the original value?

- #9

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3a^2 + 5b^2 = 17; 2ab = -4.

The second eqn establishes that a and b are some combination of +1, -2 or -1, +2.

The first settles it as a=2, b=-1 or a=-2, b=1.

a=-2, b=1 would give a negative value. By convention, real values of the √ function are taken to be positive.

Next step:

4 + √5 + 2√3 - √5 = 4 + 2√3 = (1+√3)^2

Final step:

-√3 + 1 + √3 = 1

- #10

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I found out they're called nested radicals but the stuff I found on how to solve them are way beyond me.

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