Can you help me simplify these double radicals?
1.[tex] \sqrt{6+2 \sqrt{5} } [/tex]
2.[tex] \sqrt{6+2 \sqrt{8} } [/tex]
3.[tex] \sqrt{10+2 \sqrt{21} } [/tex]
4.[tex] \sqrt{7-2 \sqrt{10} } [/tex]
5.[tex] \sqrt{10-2 \sqrt{21} } [/tex]
6.[tex] \sqrt{6-\sqrt{20} } [/tex]
7.[tex] \sqrt{7-\sqrt{48} } [/tex]
8.[tex] \sqrt{8+4\sqrt{3} } [/tex]
9.[tex] \sqrt{12-6\sqrt{3} } [/tex]
10.[tex] \sqrt{19+8\sqrt{3} } [/tex]

Respuesta :

You know that
.. (a +b)^2 = a^2 +b^2 +2ab
If these are square roots, then you have
.. (√a +√b)^2 = a +b +2√(ab)
.. (√a -√b)^2 = a +b -2√(ab)

In each case, you can put the inner radical in the form 2√something, then look for two factors of "something" that add to the other constant.
The result is then √factor1 ±√factor2, with the sign matching that under the original radical.

1. 1 +√5
2. 2 +√2
3. √3 +√7
4. √5 -√2 . . . . . . . . Technically, √(x^2) = |x|. By expressing the result as a positive number (smaller subtracted from larger), we don't have to worry about absolute value.

5. See 3 and 4. √7 -√3
6. √20 = 2√5. See 1 and 4.
.. √5 -1
7. 2 -√3
8. 4√3 = 2√12 = 2√(6*2)
.. √6 +√2
9. 6√3 = 2√(9*3)
.. 3 -√3
10. 8√3 = 2√(16*3)
.. 4 +√3
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