Conjugate (square roots)

As $${\displaystyle (a+b{\sqrt {d}})(a-b{\sqrt {d}})=a^{2}-b^{2}d}$$ 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).^[3] An example of this usage is: $${\displaystyle {\frac {a+b{\sqrt {d}}}{x+y{\sqrt {d}}}}={\frac {(a+b{\sqrt {d}})(x-y{\sqrt {d}})}{(x+y{\sqrt {d}})(x-y{\sqrt {d}})}}={\frac {ax-dby+(xb-ay){\sqrt {d}}}{x^{2}-y^{2}d}}.}$$ Hence: $${\displaystyle {\frac {1}{a+b{\sqrt {d}}}}={\frac {a-b{\sqrt {d}}}{a^{2}-db^{2}}}.}$$

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