UK Secondary (7-11)
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Conjugates and surds
Lesson

A binomial is an expression of the form $A+B$A+B, containing two terms. Changing the sign of the second term gives us the binomial $A-B$AB, which we call a conjugate for the original binomial $A+B$A+B.

If we then try to find a conjugate for the binomial $A-B$AB by changing the sign of the second term, we obtain the original binomial $A+B$A+B. That is, any binomial is a conjugate of its own conjugate. We often refer to two such binomials as a conjugate pair.

Notice that the product of a conjugate pair has a familiar form $\left(A+B\right)\left(A-B\right)$(A+B)(AB) which is the factorised form of the difference of two squares $A^2-B^2$A2B2. This observation motivates us to look at binomials containing surds - note that the expression $A^2-B^2$A2B2 will be rational even if the terms $A$A or $B$B are square roots.

 

Conjugates of binomials with surds

Consider a binomial such as $1+\sqrt{2}$1+2. We can find a conjugate for this expression in the same way - by switching the sign of the second term. Doing so, we find that $1-\sqrt{2}$12 is a conjugate for $1+\sqrt{2}$1+2.

The process is the same even if the expression is more complicated, such as $\sqrt{x}-4\sqrt{3}$x43. A conjugate for this expression would be $\sqrt{x}+4\sqrt{3}$x+43.

 

Summary

For any binomial expression $A+B$A+B, we can find a conjugate $A-B$AB by changing the sign of the second term.

A binomial and its conjugate are sometimes called a conjugate pair.

A side note

We can rewrite the binomial $A+B$A+B in the equivalent form $B+A$B+A by changing the order of the terms. By doing so we can see that $B-A$BA is also a conjugate for this expression, as well as $A-B$AB.

That is, a binomial has two possible conjugates (since there are two orders in which the binomial can be written).

 

Practice questions

Question 1

Determine a conjugate for $1+\sqrt{10}$1+10.

Question 2

Determine a conjugate for $\sqrt{5}-\sqrt{x}$5x.

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