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.

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.

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

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

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