Conjugate (square roots)
# 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.

