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:

\(\ds a\) \(=\) \(\ds c + d\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds d\) \(=\) \(\ds a - c\) subtracting $c$ from both sides


Solving for $c$:

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


Solving for $d$:

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

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

\(\ds p + q\) \(=\) \(\ds \frac a 2 + \frac {\sqrt {a^2 - b^2} } 2 + \frac a 2 - \frac {\sqrt {a^2 - b^2} } 2\)
\(\ds \) \(=\) \(\ds a\)
\(\ds \sqrt {4pq}\) \(=\) \(\ds \sqrt {4 \paren {\frac a 2 + \frac {\sqrt {a^2 - b^2} } 2} \paren {\frac a 2 - \frac {\sqrt {a^2 - b^2} } 2} }\)
\(\ds \) \(=\) \(\ds \sqrt {\paren {a + \sqrt {a^2 - b^2} } \paren {a - \sqrt {a^2 - b^2} } }\) 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:

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


$\blacksquare$