Conjugate (square roots)

In mathematics, the conjugate of an expression of the form ${\displaystyle a+b{\sqrt {d}}}$ is ${\displaystyle a-b{\sqrt {d}},}$ provided that ${\displaystyle {\sqrt {d}}}$ does not appear in a and b. One says also that the two expressions are conjugate.^[1]

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

Complex conjugation is the special case where the square root is ${\displaystyle i={\sqrt {-1}},}$ the imaginary unit.^[2]

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