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