1.06 Surds
Surds are expressions such as $\sqrt{3}$√3 or $\sqrt{39}$√39, which contain a root that can't be simplified to an integer or fraction. They are a special type of irrational number (which means they can't be written as a fraction or a ratio).
Surd rules
There are three key results that we utilise when operating with surds:
$\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$√a×b=√a×√b
$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$√ab=√a√b
$\sqrt{a}^2=\sqrt{a^2}=a$√a2=√a2=a
Simplifying surds
Sometimes an expression of the form $\sqrt{a}$√a is a surd and cannot be expressed as a rational number, but can be simplified by rewriting the larger surd using a smaller one. The key to simplifying a surd of the form $\sqrt{a}$√a is to find the highest factor of $a$a that is a square number (a number that has an integer square root). For example:
| $\sqrt{20}$√20 | $=$= | $\sqrt{4\times5}$√4×5 |
| $=$= | $\sqrt{4}\times\sqrt{5}$√4×√5 | |
| $=$= | $2\sqrt{5}$2√5 |
This is called a simplified surd.
If we had chosen to split $\sqrt{20}$√20 into $\sqrt{2\times10}=\sqrt{2}\times\sqrt{10}$√2×10=√2×√10 instead we would not have been able to simplify any further, since neither $2$2 nor $10$10 are square numbers.
Multiplication and division of surds
When using the surd rules to solve multiplication and division questions involving surds, remember to multiply or divide any numbers together in one group and the surds together in another. For example:
| $2\sqrt{2}\times3\sqrt{10}$2√2×3√10 | $=$= | $2\times3\times\sqrt{2}\times\sqrt{10}$2×3×√2×√10 |
| $=$= | $6\sqrt{20}$6√20 | |
| $=$= | $6\sqrt{4\times5}$6√4×5 | |
| $=$= | $6\times2\sqrt{5}$6×2√5 | |
| $=$= | $12\sqrt{5}$12√5 |
It is good practice to present our final answers to surd problems in fully simplified form.
Practice questions
Question 1
Simplify the expression $\sqrt{343}\times\sqrt{75}$√343×√75
Question 2
Simplify the expression $20\sqrt{14}\div5\sqrt{2}$20√14÷5√2.
Question 3
Simplify the expression $4\sqrt{35}\times\sqrt{5}$4√35×√5
Give your answer in the simplest surd form.
Addition and multiplication of surds
The rules above show that any two surds can be combined through multiplication or division. On the other hand, only "like surds" can be added. That is:
| $\sqrt{a}+\sqrt{b}$√a+√b | $\ne$≠ | $\sqrt{a+b}$√a+b |
Similarly for subtraction of surds:
| $\sqrt{a}-\sqrt{b}$√a−√b | $\ne$≠ | $\sqrt{a-b}$√a−b |
If the surds are the same value, then we have:
| $\sqrt{a}+\sqrt{a}$√a+√a | $=$= | $2\sqrt{a}$2√a |
This can be extended to adding or subtracting multiples of surds - for example:
| $5\sqrt{3}-3\sqrt{3}$5√3−3√3 | $=$= | $2\sqrt{3}$2√3 |
This is exactly the same as collecting like terms, except that we are looking for matching surds instead of pronumerals.
We can combine this with our methods of simplifying surds above, in order to rewrite surds as "like surds". For example:
| $\sqrt{12}-\sqrt{3}$√12−√3 | $=$= | $\sqrt{4\times3}-\sqrt{3}$√4×3−√3 |
| $=$= | $\sqrt{4}\times\sqrt{3}-\sqrt{3}$√4×√3−√3 | |
| $=$= | $2\sqrt{3}-\sqrt{3}$2√3−√3 | |
| $=$= | $\sqrt{3}$√3 |
So in order to determine whether two surds can be added or subtracted, it is usually best to simplify all surds first.
Practice questions
Question 4
Fully simplify the expression $5\sqrt{2}+26\sqrt{3}+22\sqrt{3}-8\sqrt{2}$5√2+26√3+22√3−8√2
Question 5
Fully simplify the expression $\sqrt{32}-\sqrt{98}$√32−√98
Question 6
Expand and simplify the expression $\left(2\sqrt{5}+3\right)^2$(2√5+3)2
Rationalising the denominator
Just as we like to present our final surd answers in fully simplified form, it is conventional to write fractions without any surds in the denominator. Another way of saying this is that we want to have a rational denominator.
If an expression contains a fraction with a surd in the denominator, we can use a technique to rewrite the fraction without any surds in the denominator. This technique is known as rationalising the denominator.
For fractions with a single term in the denominator, we can easily make the denominator rational by multiplying by the surd. To keep the fraction equivalent, we must also multiply the numerator by this same surd.
Worked example
Example 1
Rewrite the expression $\frac{8}{3\sqrt{7}}$83√7 with a rational denominator.
Think: To make the denominator $3\sqrt{7}$3√7 rational, we can multiply it by the surd part $\sqrt{7}$√7, since $\sqrt{7}\times\sqrt{7}=\sqrt{7}^2=7$√7×√7=√72=7. We will also need to multiply the numerator by $\sqrt{7}$√7 to keep the fraction equivalent.
Do:
| $\frac{8}{3\sqrt{7}}$83√7 | $=$= | $\frac{8}{3\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}$83√7×√7√7 |
| $=$= | $\frac{8\sqrt{7}}{3\sqrt{7}\times\sqrt{7}}$8√73√7×√7 | |
| $=$= | $\frac{8\sqrt{7}}{3\times7}$8√73×7 | |
| $=$= | $\frac{8\sqrt{7}}{21}$8√721 |
Example 2
Let's consider an example where we have a combination of rational and irrational terms in the denominator of a fraction such as rationalising the denominator of: $\frac{5}{\sqrt{6}-1}$5√6−1 .
We could try the same technique as before, and multiply the numerator and denominator by $\sqrt{6}$√6? Doing so, we get:
| $\frac{5}{\sqrt{6}-1}$5√6−1 | $=$= | $\frac{5}{\sqrt{6}-1}\times\frac{\sqrt{6}}{\sqrt{6}}$5√6−1×√6√6 |
| $=$= | $\frac{5\sqrt{6}}{\left(\sqrt{6}-1\right)\times\sqrt{6}}$5√6(√6−1)×√6 | |
| $=$= | $\frac{5\sqrt{6}}{\sqrt{6}\times\sqrt{6}-\sqrt{6}}$5√6√6×√6−√6 | |
| $=$= | $\frac{5\sqrt{6}}{6-\sqrt{6}}$5√66−√6 |
This new expression still has a surd term in the denominator - it isn't any simpler than when we started!
The key to solving this problem relies on the fact that squaring a surd makes it rational and the difference of two squares identity learnt in our work on quadratics. The identity has the form:
$\left(a-b\right)\left(a+b\right)=a^2-b^2$(a−b)(a+b)=a2−b2
Notice that on the left of this identity is an expression with two terms, multiplied by another expression with the same two terms separated by the opposite sign, and the result on the right has only square terms.
To use this identity, we want to multiply $\sqrt{6}-1$√6−1 by $\sqrt{6}+1$√6+1, which has the same terms but the opposite sign between them. This is known as the conjugate.
Multiplying numerator and denominator by $\sqrt{6}+1$√6+1, we get:
| $\frac{5}{\sqrt{6}-1}\times\frac{\sqrt{6}+1}{\sqrt{6}+1}$5√6−1×√6+1√6+1 | $=$= | $\frac{5\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}$5(√6+1)(√6−1)(√6+1) |
| $=$= | $\frac{5\sqrt{6}+5}{6-1}$5√6+56−1 | |
| $=$= | $\frac{5\sqrt{6}+5}{5}$5√6+55 |
We can go even further and cancel a factor of $5$5 between both numerator and denominator, which gives us the final simplified answer of:
| $\frac{5\sqrt{6}+5}{5}$5√6+55 | $=$= | $\frac{5\left(\sqrt{6}+1\right)}{5}$5(√6+1)5 |
| $=$= | $\sqrt{6}+1$√6+1 |
- To rationalise a denominator that has a single term, multiply numerator and denominator by the surd part of the denominator.
- To rationalise a denominator that has two terms, multiply numerator and denominator by the conjugate of the denominator.
Practice questions
Question 7
Express the fraction $\frac{7\sqrt{30}}{\sqrt{10}}$7√30√10 in simplest form with a rational denominator.
Question 8
Express the fraction $\frac{8}{9\sqrt{11}-5}$89√11−5 in simplest form with a rational denominator.