What is a radical conjugate?

1 Answer
May 19, 2016

Answer:

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)#