Conjugate (square roots)

In mathematics, the conjugate of an expression of the form $$a + b \sqrt d$$ is $$a - b \sqrt d,$$ provided that $$\sqrt d$$ does not appear in $a$ and $b$. One says also that the two expressions are conjugate.

In particular, the two solutions of a quadratic equation are conjugate, as per the $$\pm$$ in the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$.

Complex conjugation is the special case where the square root is $$i = \sqrt{-1},$$ the imaginary unit.

Properties
As $$(a + b \sqrt d)(a - b \sqrt d) = a^2 - b^2 d$$ and $$(a + b \sqrt d) + (a - b \sqrt d) = 2a,$$ the sum and the product of conjugate expressions do not involve the square root anymore.

This property is used for removing a square root from a denominator, by multiplying the numerator and the denominator of a fraction by the conjugate of the denominator (see Rationalisation). An example of this usage is: $$\frac{a + b \sqrt d}{x + y\sqrt d} = \frac{(a + b \sqrt d)(x - y \sqrt d)}{(x + y \sqrt d)(x - y \sqrt d)} = \frac{ax - dby + (xb - ay) \sqrt d}{x^2 - y^2 d}.$$ Hence: $$\frac{1}{a + b \sqrt d} = \frac{a - b \sqrt d}{a^2 - db^2}.$$

A corollary property is that the subtraction:
 * $$(a+b\sqrt d) - (a-b\sqrt d)= 2b\sqrt d,$$

leaves only a term containing the root.