Conjugate (square roots)

As

${\displaystyle (a+b{\sqrt {d}})(a-b{\sqrt {d}})=a^{2}-db^{2},}$

and

${\displaystyle (a+b{\sqrt {d}})+(a-b{\sqrt {d}})=2a,}$

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). Typically, one has

${\displaystyle {\frac {a_{1}+b_{1}{\sqrt {d}}}{a_{2}+b_{2}{\sqrt {d}}}}={\frac {(a_{1}+b_{1}{\sqrt {d}})(a_{2}-b_{2}{\sqrt {d}})}{(a_{2}+b_{2}{\sqrt {d}})(a_{2}-b_{2}{\sqrt {d}})}}={\frac {a_{1}a_{2}-db_{1}b_{2}+(a_{2}b_{1}-a_{1}b_{2}){\sqrt {d}}}{a_{2}^{2}-db_{2}^{2}}}.}$

In particular

${\displaystyle {\frac {1}{a+b{\sqrt {d}}}}={\frac {a-b{\sqrt {d}}}{a^{2}-db^{2}}}.}$

A corollary property is that the subtraction:

${\displaystyle (a+b{\sqrt {d}})-(a-b{\sqrt {d}})=2b{\sqrt {d}},}$

leaves only a term containing the root.

• Conjugate element (field theory), the generalization to the roots of a polynomial of any degree