# Square Root of Sum as Sum of Square Roots

## 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

$\sqrt p + \sqrt q = \sqrt {p + q + \sqrt {4pq}}$

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:

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

$\blacksquare$