Square Root of Sum as Sum of Square Roots/Proof 2
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Theorem
Let $a, b \in \R, a \ge b$.
Then:
- $\sqrt {a + b} = \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2}$
Proof
From Sum of Square Roots as Square Root of Sum:
- $\sqrt p + \sqrt q = \sqrt {p + q + \sqrt {4pq}}$
Let
- $p = \dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2$,
- $q = \dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2$.
Then
\(\ds p + q\) | \(=\) | \(\ds \frac a 2 + \frac {\sqrt {a^2 - b^2} } 2 + \frac a 2 - \frac {\sqrt {a^2 - b^2} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a\) |
\(\ds \sqrt {4pq}\) | \(=\) | \(\ds \sqrt {4 \paren {\frac a 2 + \frac {\sqrt {a^2 - b^2} } 2} \paren {\frac a 2 - \frac {\sqrt {a^2 - b^2} } 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\paren {a + \sqrt {a^2 - b^2} } \paren {a - \sqrt {a^2 - b^2} } }\) | Real Multiplication Distributes over Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {a^2 - \paren {a^2 - b^2} }\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {b^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b\) |
Therefore:
- $\sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2} = \sqrt {a + b}$
$\blacksquare$