Square Root of Number Plus Square Root

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Theorem

Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.

Then:

$\ds \sqrt {a + \sqrt b} = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}$


Proof 1

We are given that $a^2 - b > 0$.

Then:

$a > \sqrt b$

and so $\ds \sqrt {a + \sqrt b}$ is defined on the real numbers.


Let $\ds \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$ where $x, y$ are (strictly) positive real numbers.

Squaring both sides gives:

\(\ds a + \sqrt b\) \(=\) \(\ds \paren {\sqrt x + \sqrt y}^2\)
\(\ds \) \(=\) \(\ds x + y + 2 \sqrt {x y}\)


Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$




From $\sqrt b = 2 \sqrt {x y}$ we get:

\(\ds \sqrt b\) \(=\) \(\ds 2 \sqrt {x y}\)
\(\ds \leadstoandfrom \ \ \) \(\ds b\) \(=\) \(\ds 4 x y\)
\(\ds \leadstoandfrom \ \ \) \(\ds x y\) \(=\) \(\ds \frac b 4\)


By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation:

$z^2 - a z + \dfrac b 4 = 0$

From Solution to Quadratic Equation:

$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$

where $a^2 - b > 0$ (which is a given)



Without loss of generality:

\(\ds x = z_1\) \(=\) \(\ds \dfrac {a + \sqrt {a^2 - b} } 2\)
\(\ds y = z_2\) \(=\) \(\ds \dfrac {a - \sqrt {a^2 - b} } 2\)


Subsituting into $\ds \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$:

\(\ds \sqrt {a + \sqrt b}\) \(=\) \(\ds \sqrt x + \sqrt y\)
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}\)

$\blacksquare$


Proof 2

\(\ds \paren {\sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2} }^2\) \(=\) \(\ds \dfrac {a + \sqrt {a^2 - b} } 2 + \dfrac {a - \sqrt {a^2 - b} } 2 + 2 \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}\) multiplying out
\(\ds \) \(=\) \(\ds a + \sqrt {a + \sqrt {a^2 - b} } \sqrt {a - \sqrt {a^2 - b} }\) simplifying
\(\ds \) \(=\) \(\ds a + \sqrt {a^2 - \paren {a^2 - b} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds a + \sqrt b\) simplifying
\(\ds \leadsto \ \ \) \(\ds \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}\) \(=\) \(\ds \sqrt {a + \sqrt b}\) taking square root of both sides




$\blacksquare$