What is a radical conjugate?

1 Answer
May 19, 2016

Assuming that this is a maths question rather than a chemistry question, the radical conjugate of a+bsqrt(c)a+bc is a-bsqrt(c)abc

Explanation:

When simplifying a rational expression such as:

(1+sqrt(3))/(2+sqrt(3))1+32+3

we want to rationalise the denominator (2+sqrt(3))(2+3) by multiplying by the radical conjugate (2-sqrt(3))(23), formed by inverting the sign on the radical (square root) term.

So we find:

(1+sqrt(3))/(2+sqrt(3))=(1+sqrt(3))/(2+sqrt(3))*(2-sqrt(3))/(2-sqrt(3))=(sqrt(3)-1)/(4-3) = sqrt(3)-11+32+3=1+32+32323=3143=31

This is one use of the difference of squares identity:

a^2-b^2 = (a-b)(a+b)a2b2=(ab)(a+b)

Specifically:

a^2-b^2c = (a-bsqrt(c))(a+bsqrt(c))a2b2c=(abc)(a+bc)

A complex conjugate is actually a special case of the radical conjugate in which the radical is i = sqrt(-1)i=1