5.11 Rationalising the denominator (Extended)
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