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# Conjugate surds

The conjugate of the sum of two surds $\sqrt{x}\Tred{+} \sqrt{y}$ is $\sqrt{x} \Torange{-} \sqrt{y}$. We just need to change sign.

The conjugate of $1\Tred{+}\sqrt{2}$ is $1\Torange{-}\sqrt{2}$.

The conjugate of $\sqrt{3}\Tred{-} \sqrt{5}$ is $\sqrt{3}\Torange{+}\sqrt{5}$.

Conjugate surds are useful because the product is simple

$$(\Tblue{\sqrt{x}+\sqrt{y}})(\Tgreen{\sqrt{x}-\sqrt{y}}) = (\sqrt{x})^2 - (\sqrt{y})^2 = x - y.$$

Division / reciprocal of conjugate surds simplifies as well

$$\frac{1}{\Tblue{\sqrt{x} - \sqrt{y}}} = \frac{\Tgreen{\sqrt{x}+\sqrt{y}}}{(\Tblue{\sqrt{x} - \sqrt{y}})(\Tgreen{\sqrt{x}+\sqrt{y}})} = \frac{\Tgreen{\sqrt{x} +\sqrt{y}}}{x-y}.$$

Here are a few examples

\begin{align*} (\Tblue{\sqrt{2}+\sqrt{3}})(\Tgreen{\sqrt{2}-\sqrt{3}}) &= 2 - 3 = -1\\ (\Tblue{\sqrt{5}-1})(\Tgreen{\sqrt{5}+1}) & = 5-1 = 4\\ \frac{1}{\Tblue{1+\sqrt{2}}} = \frac{\Tgreen{1-\sqrt{2}}}{1-2} &= -1+\sqrt{2}\\ \frac{3}{\Tblue{\sqrt{7}-1}} = \frac{3(\Tgreen{\sqrt{7}+1})}{7-1} &= \frac{\sqrt{7}+1}{2} \end{align*}