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iGCSE (2021 Edition)

1.08 Multiplying and dividing surds

Lesson

Multiplying surds

We have seen that we can simplify surds  using the property $\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$a×b=a×b.

It follows logically that if we wanted to multiply two surds together we could combine them back in the same way. That is $\sqrt{a}\times\sqrt{b}=\sqrt{a\times b}$a×b=a×b.

If we want to find the product of more complicated surds, that may have a term outside the surd, we can use the properties of multiplication to rearrange the expression to make it easier.

Worked examples

Example 1

Simplify $2\sqrt{5}\times7\sqrt{3}$25×73.

Think: We have two terms of the form $a\sqrt{b}$ab, which is the same as $a\times\sqrt{b}$a×b. So we are really finding the product of four terms: $2$2, $\sqrt{5}$5, $7$7 and $\sqrt{3}$3.

This means we can rearrange the terms in any order without changing the result. We can make this much easier by finding the product of the two integer terms and the product of the roots.

Do: Rearranging the equation gives us:

$\left(2\times7\right)\times\left(\sqrt{5}\times\sqrt{3}\right)=14\times\sqrt{5\times3}$(2×7)×(5×3)=14×5×3

Which simplifies to $14\sqrt{15}$1415.

As $15$15 does not have any square factors we cannot simplify it further.

 

Example 2

Simplify $9\sqrt{28}\times3\sqrt{7}$928×37.

Think: As before we have two terms of the form $a\sqrt{b}$ab, so we will rearrange the expression in the same way.

Do: Rearranging the equation gives us:

$\left(9\times3\right)\times\left(\sqrt{28}\times\sqrt{7}\right)=27\times\sqrt{28\times7}$(9×3)×(28×7)=27×28×7

Which simplifies to $27\sqrt{196}$27196.

Now, in this case, if we get the prime factorisation of $196$196 we find $196=2^2\times7^2$196=22×72. So it is the product of two square numbers, which means it is itself a square number as $a^2\times b^2=\left(a\times b\right)^2$a2×b2=(a×b)2.

In fact, $196=2^2\times7^2$196=22×72$=$=$14^2$142, which means:

$27\times\sqrt{196}=27\times14$27×196=27×14$=$=$378$378

 

So our original expression with two surds ends up simplifying to an integer.

Reflect: We could have approached this in a different manner by noticing that one of the terms in the original expression $\sqrt{28}$28 can be simplified straight away:

$\sqrt{28}$28 $=$= $\sqrt{4\times7}$4×7
  $=$= $\sqrt{4}\times\sqrt{7}$4×7
  $=$= $2\sqrt{7}$27

 

This means we can rewrite the original expression:

$9\sqrt{28}\times3\sqrt{7}=9\times2\sqrt{7}\times3\sqrt{7}$928×37=9×27×37

We can now rearrange the equation:

$\left(9\times2\times3\right)\times\left(\sqrt{7}\times\sqrt{7}\right)$(9×2×3)×(7×7)

Notice that we now have $\sqrt{7}\times\sqrt{7}$7×7 which we know is just equal to $7$7. This means we now just have the product of four integers.

$9\times2\times3\times7=378$9×2×3×7=378

Which is the same answer we arrived at before. So sometimes it may be easier to try and simplify the surds before trying to multiply them together but, rest assured, the answer is the same which ever method is used!

 

Dividing surds

It follows that we can treat the division of surds in the same way. That is $\sqrt{a}\div\sqrt{b}=\sqrt{a\div b}$a÷​b=a÷b. We can also split terms up so that we can evaluate the quotient of the integer terms, and the surd terms separately.

Worked example

Example 3

Simplify the expression $\frac{4\sqrt{18}}{2\sqrt{3}}$41823.

Think: We want to split the surd parts and integer parts up to make it easier.

If we use the rule $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$ab=ab, we can then simplify the fraction under the square root symbol.

Do: 

$\frac{4\sqrt{18}}{2\sqrt{3}}$41823 $=$= $\frac{4}{2}\times\frac{\sqrt{18}}{\sqrt{3}}$42×183

Separating division into surd and integer parts

  $=$= $\frac{4}{2}\times\sqrt{\frac{18}{3}}$42×183

Simplifying surds into one surd

  $=$= $2\sqrt{6}$26

Simplifying the two quotients

 

Reflect: Similar to the multiplication example above, we can approach this question by first trying to simplify surds.

$\sqrt{18}=\sqrt{9\times2}$18=9×2$=$=$3\sqrt{2}$32

This means we can rewrite the original expression:

$\frac{4\sqrt{18}}{2\sqrt{3}}$41823 $=$= $\frac{4\times3\sqrt{2}}{2\sqrt{3}}$4×3223

Simplifying $\sqrt{18}$18

  $=$= $\frac{12}{2}\times\frac{\sqrt{2}}{\sqrt{3}}$122×23

Separating division into surd and integer parts

  $=$= $\frac{12}{2}\times\sqrt{\frac{2}{3}}$122×23

Simplifying surds into one surd

  $=$= $6\sqrt{\frac{2}{3}}$623

Simplifying the integer quotient

 

Now, this appears to be a different answer to the one we found above, but using a 'trick' and our knowledge of equivalent fractions, we can rewrite this to equal $2\sqrt{6}$26.

$\sqrt{\frac{2}{3}}=\sqrt{\frac{6}{9}}$23=69

Rewriting $\sqrt{\frac{2}{3}}$23 like this means we now have a square term in the denominator. We can now split this surd up using the rule $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ab=ab.

$\sqrt{\frac{6}{9}}=\frac{\sqrt{6}}{\sqrt{9}}$69=69$=$=$\frac{\sqrt{6}}{3}$63

So, we can now rewrite the final answer of $6\sqrt{\frac{2}{3}}$623 to remove the fraction inside the root sign.

$6\sqrt{\frac{2}{3}}$623 $=$= $6\times\frac{\sqrt{6}}{3}$6×63
  $=$= $\frac{6}{3}\times\sqrt{6}$63×6
  $=$= $2\sqrt{6}$26

 

So unlike the multiplication example where simplifying first lead to an easier calculation, in this case it ended up leading to more steps, but we still arrived at the same final answer. Which approach is best is often up to the numbers involved in the question.

 

Multiplying and dividing surds

To multiply terms involving surds use the rule:

$a\sqrt{m}\times b\sqrt{n}=ab\times\sqrt{mn}$am×bn=ab×mn

To divide terms involving surds use the rule:

$a\sqrt{m}\div b\sqrt{n}=\frac{a}{b}\times\sqrt{\frac{m}{n}}$am÷​bn=ab×mn

 

Practice questions

Question 1

Express $\sqrt{19}\times\sqrt{17}$19×17 as a single surd. That is, in the form $\sqrt{n}$n.

Question 2

Simplify: $8\sqrt{15}\times8\sqrt{5}$815×85

  1. Give your answer in simplest surd form.

Question 3

Simplify: $\frac{\sqrt{51}}{\sqrt{17}}$5117

Question 4

Expand and simplify: $3\sqrt{3}\left(\sqrt{13}-5\right)$33(135)

Outcomes

0606C4.1B

Perform simple operations with surds.

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