2.05 Rationalising the denominator (Enrichment)
Rationalising the denominator is an expression that means to rewrite a fraction that has surds in the denominator in such a way that we only have an integer in the denominator. By convention this is the preferred way to write fractions involving surds; it makes adding fractions involving surds much more simple and in the time before calculators it made calculations by hand much easier. So how can we achieve this>?
Using perfect squares
We know that when we square a surd, the answer is always going to be rational. This is the key we need to rationalise fractions such as $\frac{8}{3\sqrt{7}}$83√7.
What do you think will happen if we multiply the denominator here by $\sqrt{7}$√7?
Well, the square root sign will disappear, but we also need to make sure the fraction still has the same value, so let's try multiplying the fraction by $\frac{\sqrt{7}}{\sqrt{7}}=1$√7√7=1, which will not change the value of the fraction at all.
$\frac{8}{3\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}=\frac{8\times\sqrt{7}}{3\times\sqrt{7}\times\sqrt{7}}$83√7×√7√7=8×√73×√7×√7
which simplifies to
$\frac{8\sqrt{7}}{3\times\left(\sqrt{7}\right)^2}=\frac{8\sqrt{7}}{3\times7}$8√73×(√7)2=8√73×7
This answer can again be finally simplified to: $\frac{8\sqrt{7}}{21}$8√721
The fraction now looks completely different! But try putting it in your calculator, do you get the same value as $\frac{8}{3\sqrt{7}}$83√7?
So now we know one way of rationalising the denominator is to multiply top and bottom by the surd in the denominator.
To rationalise the denominator of the form $\frac{a}{b\sqrt{n}}$ab√n we want to multiply it by the fraction $\frac{\sqrt{n}}{\sqrt{n}}$√n√n:
$\frac{a}{b\sqrt{n}}\times\frac{\sqrt{n}}{\sqrt{n}}=\frac{a\sqrt{n}}{bn}$ab√n×√n√n=a√nbn
Using differences of two squares
The above technique is great for when there is only one term in the denominator, but it doesn't work when we have a binomial expression in the denominator.
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 $A-B$A−B, which we call a conjugate for the original binomial $A+B$A+B.
If we then try to find a conjugate for the binomial $A-B$A−B 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)(A−B) which is the factorised form of the difference of two squares $A^2-B^2$A2−B2. This means the expression $A^2-B^2$A2−B2 will always be rational even if the terms $A$A or $B$B are surds.
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}$1−√2 is a conjugate for $1+\sqrt{2}$1+√2.
The process is the same even if the expression is more complicated, such as $\sqrt{x}-4\sqrt{3}$√x−4√3. A conjugate for this expression would be $\sqrt{x}+4\sqrt{3}$√x+4√3.
For any binomial expression $A+B$A+B, we can find a conjugate $A-B$A−B by changing the sign of the second term.
A binomial and its conjugate are sometimes called a conjugate pair.
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 $B-A$B−A is also a conjugate for this expression, as well as $A-B$A−B.
That is, a binomial has two possible conjugates (since there are two orders in which the binomial can be written).
Worked example
Rationalise the denominator of $\frac{5}{\sqrt{6}-1}$5√6−1.
Think: If we use the perfect squares technique and multiply the bottom by $\sqrt{6}$√6, it'll become $6-\sqrt{6}$6−√6. So whilst it got rid of the original $\sqrt{6}$√6, it has turned the $1$1 into a $\sqrt{6}$√6.
We want to multiply the numerator and denominator by the conjugate of the denominator $\sqrt{6}-1$√6−1. We want to change the sign of this expression to find the conjugate, that is, $\sqrt{6}+1$√6+1, so let's multiply the expression by $\frac{\sqrt{6}+1}{\sqrt{6}+1}$√6+1√6+1.
Do:
| $\frac{5}{\sqrt{6}-1}$5√6−1 | $=$= | $\frac{5}{\sqrt{6}-1}\times\frac{\sqrt{6}+1}{\sqrt{6}+1}$5√6−1×√6+1√6+1 |
Multiplying by the conjugate of the denominator |
| $=$= | $\frac{5\times\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\times\left(\sqrt{6}+1\right)}$5×(√6+1)(√6−1)×(√6+1) |
Combinging the fractions |
|
| $=$= | $\frac{5\sqrt{6}+5}{\left(\sqrt{6}\right)^2+\sqrt{6}-\sqrt{6}-1}$5√6+5(√6)2+√6−√6−1 |
Expanding the brackets |
|
| $=$= | $\frac{5\sqrt{6}+5}{6-1}$5√6+56−1 |
Simplifying the expression in the denominator |
|
| $=$= | $\frac{5\sqrt{6}+5}{5}$5√6+55 |
Evaluating the difference in the denominator |
|
| $=$= | $\sqrt{6}+1$√6+1 |
Cancelling common factors |
Reflect: By rationalising the denominator we have in fact gotten rid of the denominator entirely, and we are left with the much simpler expression of $\sqrt{6}=1$√6=1.
To rationalise the denominator of the form $\frac{a}{b\sqrt{m}+c\sqrt{n}}$ab√m+c√n we want to multiply it by the fraction $\frac{b\sqrt{m}-c\sqrt{n}}{b\sqrt{m}-c\sqrt{n}}$b√m−c√nb√m−c√n:
$\frac{a}{b\sqrt{m}+c\sqrt{n}}\times\frac{b\sqrt{m}-c\sqrt{n}}{b\sqrt{m}-c\sqrt{n}}=\frac{a\left(b\sqrt{m}-c\sqrt{n}\right)}{b^2m-c^2n}$ab√m+c√n×b√m−c√nb√m−c√n=a(b√m−c√n)b2m−c2n
Practice questions
Question 1
Express the following fraction in simplest surd form with a rational denominator::
$\frac{2}{\sqrt{6}}$2√6
Question 2
Simplify $\frac{\sqrt{39}+\sqrt{6}}{\sqrt{3}}$√39+√6√3.
Question 3
Express the following fraction in simplest surd form with a rational denominator:
$\frac{3}{5\sqrt{2}-4}$35√2−4