What is the radical conjugate of #12-sqrt(5)# ?

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Answer 1 · 2016-06-25T06:30:26

#12+sqrt(5)#

The conjugate of #12-sqrt(5)# is #12+sqrt(5)#.

This has the property that: #(12-sqrt(5))(12+sqrt(5))# is rational:

#(12-sqrt(5))(12+sqrt(5)) = 12^2-(sqrt(5))^2 = 144-5 = 139#

Typical examples where you would use the conjugate would be:

When rationalising the denominator of a quotient.
When looking at zeros of a polynomial with rational (typically integer) coefficients.

For example:

#(2+3sqrt(5))/(12-sqrt(5))#

#=((2+3sqrt(5))(12+sqrt(5)))/((12-sqrt(5))(12+sqrt(5)))#

#=(24+2sqrt(5)+36sqrt(5)+15)/(144-5)#

#=(39+38sqrt(5))/139#

The simplest polynomial with rational coefficients and zero #12-sqrt(5)# is:

#(x-(12-sqrt(5)))(x-(12+sqrt(5)))=x^2-24x+139#