Question

What is a radical conjugate?

Answer 1

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

Explanation:

When simplifying a rational expression such as:

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

we want to rationalise the denominator `(2+sqrt(3))` by multiplying by the radical conjugate `(2-sqrt(3))`, 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)-1`

This is one use of the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

Specifically:

`a^2-b^2c = (a-bsqrt(c))(a+bsqrt(c))`

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