Conjugate (square roots)

Template:Refimprove Template:Inline Template:Short description Template:About 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 Template:Mvar and Template:Mvar. 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: $$\frac{1}{a+b\sqrt{d}}=\frac{a-b\sqrt{d}}{a^2-d{b}^2}.$$

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.

See also

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