Simplifying Surds – A Level Maths Revision

Simplifying Surds - AS Level

As you make the vast step up to advanced study in maths, you will need a firm grasp on the fundamentals of simplifying surds.

Before you dive right into the nitty-gritty of worked examples, Beyond: Advanced can offer you the following resources (great for teachers and students, alike).

  • For questions and answers on this topic, click here.
  • For a surds treasure hunt, with answers, click here.

Simplifying Surds – Background

The name β€œsurd” comes from the Latin surdus, which translates as β€œmute” or β€œunspoken”. This used to apply to all irrational numbers but, now, it refers to a specific type of irrational number: those that can be written as the square root of an integer.

You will have come across Surds at GCSE; at A level it is important that you understand and can manipulate surds.

To simplify surds, there are two rules that you need to be able to use:

\sqrt{ab}=\sqrt{a} \times \sqrt{b}


Simplifying Surds – Example Content

Example Question 1:

Simplify the surd \sqrt{75}.

When simplifying surds, we need to find a factor that is a square number. In this case, we can see that 25 is a factor of 75, so we can write:

\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25}\sqrt{3} = 5\sqrt{3}

Example Question 2:

Simplify the surd \sqrt{\frac {7}{81}}  .

The square root of a fraction can be calculated by finding the square root of both the denominator and numerator:

\sqrt{\frac{7}{81}} = \frac{\sqrt{7}}{\sqrt{81}} = \frac{\sqrt{7}}{9}

Example Question 3:

Find (3\sqrt{7} - 4\sqrt{6})^2  . Give your answer as a surd in its simplest form.

It is important to remember that (a + b)^2 = (a + b)(a + b)  . We can expand brackets that include surds in exactly the same way as we would expand algebraic brackets:

(3\sqrt{7} - 4\sqrt{6})^2 = (3\sqrt{7} - 4\sqrt{6})(3\sqrt{7} - 4\sqrt{6})

= (3\sqrt{7})^2 - 2 \times (4\sqrt{6}) \times (3\sqrt{7}) + (4\sqrt{6})^2

= 63 - 24\sqrt{42} + 96

= 159 - 24\sqrt{42}

Example Question 4:

Express \sqrt{294} - \sqrt{150} in the form k\sqrt{x} where k and x are integers.

It is hard to see how to simplify \sqrt{294} , so let’s start by looking at \sqrt{150} . You may be able to spot that 150 is a multiple of 25, so we can write:

\sqrt{150} = \sqrt{6 \times 25} = 5\sqrt{6}

This then helps us simplify \sqrt{294} as we can assume that 6 is a factor:

\sqrt{294} = \sqrt{6 \times 49} = 7\sqrt{6}

Now, we can simplify \sqrt{294} - \sqrt{150} :

\sqrt{294} - \sqrt{150} = 7\sqrt{6} - 5\sqrt{6} = 2\sqrt{6}

When a surd is written on the denominator of a fraction, we rationalise the denominator to make it simpler. We do this by multiplying the numerator and denominator by an expression that will simplify the surd on the denominator.

Example Question 5:

Simplify \frac{5}{\sqrt{32}} .

In this case, we start by simplifying \sqrt{32} :

\sqrt{32}=\sqrt{16 \times 2} = 4\sqrt{2}

We now have:

\frac{5}{\sqrt{32}} = \frac{5}{4\sqrt{2}}

We can rationalise the denominator by multiplying the numerator and denominator by \sqrt{2} .

We can do this because \sqrt{2} \times \sqrt{2} = 2 , which removes the root, and \frac{\sqrt{2}}{\sqrt{2}}=1    , so our multiplication does not affect the value of the fraction.

\frac{5}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{4\sqrt{2} \times \sqrt{2}}=\frac{5\sqrt{2}}{8}

We could have rationalised this surd by multiplying the numerator and denominator by \sqrt{32} . However, by simplifying the surd first, we make sure our fraction is simplified and make the multiplication easier.

Example Question 6:

Simplify \frac{4}{5 \: - \: 2\sqrt{3}} .

To do this, we consider 5 + 2\sqrt{3} . This is called the conjugate of 5 - 2\sqrt{3} ; it can be found by simply changing the sign of the surd. To help understand how this works, think back to the difference of two squares:

(a+b)(a-b)=a^2 - b^2

When we expand (5 + 2\sqrt{3})(5 - 2\sqrt{3}) , we get:

(5 + 2\sqrt{3})(5 - 2\sqrt{3}) = 5^2 - 5 \times 2\sqrt{3} + 5 \times 2\sqrt{3} - (2\sqrt{3})^2

= 25 - 4 \times 3

= 13

We can now use this to remove the surd on the denominator:

\frac{4}{5  \:  -  \:  2\sqrt{3}} \times \frac{5  \:  + \:  2\sqrt{3}}{5  \:  +  \:  2\sqrt{3}} = \frac{4(5 \:  +  \:  2\sqrt{3})}{(5 \: - \: 2\sqrt{3})(5 \: + \: 2\sqrt{3})}

= \frac{20 \: + \: 8\sqrt{3}}{13}

So, to rationalise the denominator of a surd, we multiply the numerator and denominator by the conjugate of the denominator.

Example Question 7:

Simplify \frac{4  \: + \: \sqrt{7}}{9 \: + \: 2\sqrt{7}} .

This time, our conjugate is 9 - 2\sqrt{7} ; we multiply our numerator and denominator by this. Since the numerator is also a surd, you may find it helpful to multiply them separately:

(4+\sqrt{7})(9-2\sqrt{7}) = 36 + 9\sqrt{7} - 8\sqrt{7} - 14

= 22 + \sqrt{7}

(9 + 2\sqrt{7})(9 - 2\sqrt{7}) = 81 + 18\sqrt{7} - 18\sqrt{7} - 28

= 53

Bringing these together gives:

\frac{4  \: + \: \sqrt{7}}{9 \: + \: 2\sqrt{7}} =  \frac{4  \: + \: \sqrt{7}}{9 \: + \: 2\sqrt{7}} \times \frac{9 \: - \: 2\sqrt{7}}{9 \: - \: 2\sqrt{7}}

= \frac{22 \: + \: \sqrt{7}}{53}

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