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A Problem on Ordinary Nested Radicals

Problem

Prove or disprove that
\sqrt {7+\sqrt{33}} + \sqrt {6+\sqrt{35} } = \sqrt {5+\sqrt{21}} + \sqrt {8+\sqrt{55}}

Solution

In order to simplify the radicals, the radicands should be forced to equal square numbers (e.g., 7+\sqrt{33} should be a square of some number). Numbers whose squares have a rational and radical part are usually in the form x+y .
So let \sqrt{7 +\sqrt{33}} = x+y = \sqrt{(x+y)^2} = \sqrt {x^2+y^2+2xy}
and set
x^2+y^2=7 and 2xy=\sqrt{33} \, , i.e., y=\sqrt{33}/2x
Thus x^2 + \left ( \frac {\sqrt{33}}{2x} \right )^2 =7
which on simplification yields x=\sqrt{22}/2
And also y=\sqrt{6}/2

Thus,
\mathbf {\sqrt{7+\sqrt{33}} = \frac {\sqrt{22}+\sqrt{6}}{2} }
Using the same process for other radicals:
\mathbf {\sqrt{6+\sqrt{35}} = \frac{\sqrt{10}+\sqrt{14}}{2} }
\mathbf {\sqrt{8+\sqrt{55}} = \frac{\sqrt{10}+\sqrt{22}}{2} }
\mathbf {\sqrt{5+\sqrt{21}} = \frac{\sqrt{6}+\sqrt{14}}{2} }
Thus, now we can easily prove (by addition) that

\sqrt {7+\sqrt{33}} + \sqrt {6+\sqrt{35}} =\sqrt {5+\sqrt{21}} + \sqrt {8+\sqrt{55}}

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By in Math, Problems on .

1 Comment

  1. quentin bennett says:
    Thursday, March 17th, 2011 at 12:33

    Thxs, this was good & sorely needed practice!

    Reply

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