Square Root of Sum as Sum of Square Roots

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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 1

Let $\sqrt {a + b}$ be expressed in the form $\sqrt c + \sqrt d$.

From Square of Sum:

$a + b = c + d + 2 \sqrt {c d}$

We now need to solve the simultaneous equations:

$a = c + d$
$b = 2 \sqrt {c d}$


First:

\(\displaystyle a\) \(=\) \(\displaystyle c + d\)
\((1):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle d\) \(=\) \(\displaystyle a - c\) subtracting $c$ from both sides


Solving for $c$:

\(\displaystyle b\) \(=\) \(\displaystyle 2 \sqrt {c d}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle b^2\) \(=\) \(\displaystyle 4 c d\) squaring both sides
\(\displaystyle \) \(=\) \(\displaystyle 4 c \left({a - c}\right)\) substituting $d = a - c$ from $(1)$
\(\displaystyle \) \(=\) \(\displaystyle 4 a c - 4 c^2\) Real Multiplication Distributes over Addition
\(\displaystyle \leadsto \ \ \) \(\displaystyle 4 c^2 - 4 a c + b^2\) \(=\) \(\displaystyle 0\) adding $4 c^2 - 4 a c$ to both sides
\((2):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle c\) \(=\) \(\displaystyle \frac a 2 \pm \frac {\sqrt {a^2 - b^2} } 2\) Quadratic Formula


Solving for $d$:

\(\displaystyle d\) \(=\) \(\displaystyle a - c\)
\(\displaystyle \) \(=\) \(\displaystyle a - \frac a 2 \mp \frac {\sqrt {a^2 - b^2} } 2\) substituting $c = \dfrac a 2 \pm \dfrac {\sqrt {a^2 - b^2} } 2$ from $(2)$
\(\displaystyle \) \(=\) \(\displaystyle \frac a 2 \mp \frac {\sqrt {a^2 - b^2} } 2\)

From Real Addition is Commutative, the sign of the square root may be chosen arbitrarily, provided opposite signs are chosen for $c$ and $d$.

$\blacksquare$


Proof 2

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

\(\displaystyle p + q\) \(=\) \(\displaystyle \frac a 2 + \frac {\sqrt {a^2 - b^2} } 2 + \frac a 2 - \frac {\sqrt {a^2 - b^2} } 2\)
\(\displaystyle \) \(=\) \(\displaystyle a\)
\(\displaystyle \sqrt {4pq}\) \(=\) \(\displaystyle \sqrt {4 \left({\frac a 2 + \frac {\sqrt {a^2 - b^2} } 2}\right) \left({\frac a 2 - \frac {\sqrt {a^2 - b^2} } 2}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\left({a + \sqrt {a^2 - b^2} }\right) \left({a - \sqrt {a^2 - b^2} }\right)}\) Real Multiplication Distributes over Addition
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {a^2 - \left({a^2 - b^2}\right)}\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {b^2}\)
\(\displaystyle \) \(=\) \(\displaystyle b\)

Therefore:

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


$\blacksquare$