📂Lemmas

Square Root Expansion Formula

Formula

When $a \gt b$,

$$\sqrt{a + b \pm 2\sqrt{a b}} = \sqrt{a} \pm \sqrt{b}$$

Explanation

It might seem incredibly difficult to solve with two roots, but it can be directly solved in a perfect square form.

Example

$$\begin{align*} \sqrt{13 - 2\sqrt{12}} &= \sqrt{12 + 1 - 2\sqrt{12 \cdot 1}} \\ &= \sqrt{(\sqrt{12})^{2} + (\sqrt{1})^{2} - 2\sqrt{12}\sqrt{1}} \\ &= \sqrt{(\sqrt{12} - \sqrt{1})^{2}} \\ &= \sqrt{12} - \sqrt{1} \\ &= 2\sqrt{3} - 1 \\ \end{align*}$$

Proof

$$\begin{align*} \sqrt{a + b \pm 2\sqrt{a b}} &= \sqrt{(\sqrt{a})^{2} + (\sqrt{b})^{2} \pm 2\sqrt{a}\sqrt{b}} \\ &= \sqrt{(\sqrt{a} \pm \sqrt{b})^{2}} \\ &= \sqrt{a} \pm \sqrt{b} \end{align*}$$

In the second equality, the multiplication formula $(a \pm b)^{2} = a^{2} + b^{2} \pm 2 ab$ was used.

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