Square Root of Number Plus or Minus Square Root
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Theorem
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
Square Root of Number Plus Square Root
- $\ds \sqrt {a + \sqrt b} = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}$
Square Root of Number Minus Square Root
- $\ds \sqrt {a - \sqrt b} = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}$
Examples
\(\ds \sqrt {2 - \sqrt 3}\) | \(=\) | \(\ds \sqrt {\dfrac {2 + \sqrt {2^2 - 3} } 2} - \sqrt {\dfrac {2 - \sqrt {2^2 - 3} } 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\dfrac 3 2} - \sqrt {\dfrac 1 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\sqrt 3 - 1} {\sqrt 2}\) |
\(\ds \sqrt {3 - \sqrt 8}\) | \(=\) | \(\ds \sqrt {\dfrac {3 + \sqrt {3^2 - 8} } 2} - \sqrt {\dfrac {3 - \sqrt {3^2 - 8} } 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\dfrac 4 2} - \sqrt {\dfrac 2 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt 2 - 1\) |