What is the conjugate of #sqrt((a+b))+7# ?

1 Answer
Jan 29, 2017

Answer:

You can use #-sqrt((a+b))+7# or #sqrt((a+b))-7# as the conjugate of #sqrt((a+b))+7#

Explanation:

It actually does not matter. You can use #-sqrt((a+b))+7# as the conjugate, or #sqrt((a+b))-7#.

By convention, if a binomial expression contains a rational and an irrational term, then we usually put the rational term first, e.g.:

#2+3sqrt(5)#

Then the radical conjugate is usually taken to be:

#2-3sqrt(5)#

i.e. reversing the sign of the irrational term.

Note that that does not cover the case of:

#sqrt(2)+sqrt(3)#

for which we could use the following expression as a conjugate:

#sqrt(2)-sqrt(3)#

Alternatively, we could use the following expression as a conjugate:

#sqrt(3)-sqrt(2)#

The fundamental idea is that a conjugate is an expression which when multiplied by the original results in a rational result.

For binomials with terms that involve rationals and square roots (but not square roots of square roots), you can form a conjugate by reversing the sign of either term.

This works because of the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

So by reversing the sign of one term, we end up with a product only involving squares of the original terms.