If you know that #3 + sqrt(11)# is a root of a polynomial function, then the name given to #3 - sqrt(11)# , another root of the same function , is a __ conjugate. ?

1 Answer
George C.
Dec 17, 2017

radical conjugate

Explanation:

A conjugate is an object which when combined with the original object makes some kind of whole.

Note that the difference of squares identity tells us that:

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

So if #a# and #b# are terms consisting of square roots (possibly including #i# which is a square root of #-1#), then we can simplify #(a+b)# by multiplying by the conjugate #(a-b)# (or vice versa).

In the case of square roots, this is a radical conjugate.

For example, we find:

#(x-(3-sqrt(11)))(x-(3+sqrt(11))) = ((x-3)-sqrt(11))((x-3)+sqrt(11))#

#color(white)((x-(3-sqrt(11)))(x-(3+sqrt(11)))) = (x-3)^2-(sqrt(11))^2#

#color(white)((x-(3-sqrt(11)))(x-(3+sqrt(11)))) = x^2-6x+9-11#

#color(white)((x-(3-sqrt(11)))(x-(3+sqrt(11)))) = x^2-6x-2#

Note that the resulting quadratic polynomial has only rational coefficients - we have successfully eliminated the irrational #sqrt(11)# by using the radical conjugate.

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