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# 1.5 Estimating surds

## 1.5 Estimating surds (EMA9)

If the $$n^{\text{th}}$$ root of a number cannot be simplified to a rational number, we call it a surd. For example, $$\sqrt{2}$$ and $$\sqrt{6}$$ are surds, but $$\sqrt{4}$$ is not a surd because it can be simplified to the rational number $$\text{2}$$.

In this chapter we will look at surds of the form $$\sqrt[n]{a}$$ where $$a$$ is any positive number, for example, $$\sqrt{7}$$ or $$\sqrt{5}$$. It is very common for $$n$$ to be $$\text{2,}$$ so we usually do not write $$\sqrt{a}$$. Instead we write the surd as just $$\sqrt{a}$$

It is sometimes useful to know the approximate value of a surd without having to use a calculator. For example, we want to be able to estimate where a surd like $$\sqrt{3}$$ is on the number line. From a calculator we know that $$\sqrt{3}$$ is equal to $$\text{1,73205...}$$. It is easy to see that $$\sqrt{3}$$ is above $$\text{1}$$ and below $$\text{2}$$. But to see this for other surds like $$\sqrt{18}$$, without using a calculator you must first understand the following:

If $$a$$ and $$b$$ are positive whole numbers, and $$a<b$$, then $$\sqrt[n]{a}<\sqrt[n]{b}$$

A perfect square is the number obtained when an integer is squared. For example, $$\text{9}$$ is a perfect square since $${3}^{2}=9$$.

Similarly, a perfect cube is a number which is the cube of an integer. For example, $$\text{27}$$ is a perfect cube, because $${3}^{3}=27$$.

Consider the surd $$\sqrt{52}$$. It lies somewhere between $$\text{3}$$ and $$\text{4,}$$ because $$\sqrt{27}=3$$ and $$\sqrt{64}=4$$ and $$\text{52}$$ is between $$\text{27}$$ and $$\text{64}$$.

The following video explains how to estimate a surd.

Video: 2DDV

## Worked example 5: Estimating surds

Find two consecutive integers such that $$\sqrt{26}$$ lies between them. (Remember that consecutive integers are two integers that follow one another on the number line, for example, $$\text{5}$$ and $$\text{6}$$ or $$\text{8}$$ and $$\text{9}$$.)

### Use perfect squares to estimate the lower integer

$${5}^{2}=25$$. Therefore $$5<\sqrt{26}$$.

### Use perfect squares to estimate the upper integer

$${6}^{2}=36$$. Therefore $$\sqrt{26}<6$$.

$$5<\sqrt{26}<6$$

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## Worked example 6: Estimating surds

Find two consecutive integers such that $$\sqrt{49}$$ lies between them.

### Use perfect cubes to estimate the lower integer

$${3}^{3}=27$$, therefore $$3<\sqrt{49}$$.

### Use perfect cubes to estimate the upper integer

$${4}^{3}=64$$, therefore $$\sqrt{49}<4$$.

$$3<\sqrt{49}<4$$

### Check the answer by cubing all terms in the inequality and then simplify

$$27<49<64$$. This is true, so $$\sqrt{49}$$ lies between $$\text{3}$$ and $$\text{4}$$.

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Textbook Exercise 1.3

Determine between which two consecutive integers the following numbers lie, without using a calculator:

$$\sqrt{18}$$

$4 \text{ and } 5 ~(4^2 = 16 \text{ and } 5^2=25)$

$$\sqrt{29}$$

$5 \text{ and } 6 ~(5^2 = 25 \text{ and } 6^2=36)$

$$\sqrt{5}$$

$1 \text{ and } 2 ~(1^3 = 1 \text{ and } 2^3=8)$

$$\sqrt{79}$$

$4 \text{ and } 5~ (4^3 = 64 \text{ and } 5^3=125)$

$$\sqrt{155}$$

$12 \text{ and } 13~ (12^2 = 144 \text{ and } 13^2=169)$

$$\sqrt{57}$$

$7 \text{ and } 8~ (7^2 = 49 \text{ and } 8^2 = 64)$

$$\sqrt{71}$$

$8 \text{ and } 9~ (8^2 = 64 \text{ and } 9^2 = 81)$
$$\sqrt{123}$$
$4 \text{ and } 5~ (4^3 = 64 \text{ and } 5^3=125)$
$$\sqrt{90}$$
$4 \text{ and } 5~ (4^3 = 64 \text{ and } 5^3=125)$
$$\sqrt{81}$$
$4 \text{ and } 5~ (4^3 = 64 \text{ and } 5^3=125)$

Estimate the following surds to the nearest $$\text{1}$$ decimal place, without using a calculator.

$$\sqrt{10}$$

Since $$3^2 = 9 \text{ and } 4^2 = 16$$, $$\sqrt{10}$$ must lie between 3 and 4. But we note that 10 is closer to 9 than to 16 and so $$\sqrt{10}$$ will be closer to 3 than to 4.

$$\text{3,1}$$ or $$\text{3,2}$$ are suitable estimates.

$$\sqrt{82}$$

Since $$9^2 = 81 \text{ and } 10^2 = 100$$, $$\sqrt{82}$$ must lie between 9 and 10. But we note that 82 is closer to 81 than to 100 and so $$\sqrt{82}$$ will be closer to 9 than to 10.

$$\text{9,1}$$ is a suitable estimate.

$$\sqrt{15}$$

Since $$3^2 = 9 \text{ and } 4^2 = 16$$, $$\sqrt{15}$$ must lie between 3 and 4. But we note that 15 is closer to 16 than to 9 and so $$\sqrt{15}$$ will be closer to 4 than to 3.

$$\text{3,9}$$ is a suitable estimate.

$$\sqrt{90}$$

Since $$9^2 = 81 \text{ and } 10^2 = 100$$, $$\sqrt{90}$$ must lie between 9 and 10. But we note that 90 is about halfway between 81 and 100, so $$\sqrt{90}$$ will be halfway between 3 and 4.

$$\text{3,5}$$ is a suitable estimate.

Consider the following list of numbers:

$$\frac{27}{7} \; ; \; \sqrt{19} \; ; \; 2\pi \; ; \; \text{0,45} \; ; \; \text{0,}\overline{45} \; ; \; -\sqrt{\frac{9}{4}} \; ; \; 6 \; ; \; -\sqrt{8} \; ; \; \sqrt{51}$$

Without using a calculator, rank all the numbers in ascending order.

Remember that negative numbers are smaller than positive numbers. It may also be helpful to write the fractions as decimals to help you estimate the number. For the surds you can estimate between which two numbers the surd lies and use that to help you rank these numbers.

• $$\frac{27}{7} \approx \text{3,857}$$
• $$\sqrt{19}$$ lies between 4 and 5
• $$2\pi \approx \text{6,28}$$
• $$-\sqrt{\frac{9}{4}} = -\frac{3}{2} = -\text{1,5}$$
• $$-\sqrt{8}$$ lies between $$-\text{2}$$ and $$-\text{3}$$
• $$\sqrt{51}$$ lies between 7 and 8

Also note that $$\text{0,45} < \text{0,}\overline{45}$$.

Therefore we get the following order: $$-\sqrt{8} \; ; \; -\sqrt{\frac{9}{4}} \; ; \; \text{0,45} \; ; \; \text{0,}\overline{45} \; ; \; \frac{27}{7} \; ; \; \sqrt{19} \; ; \; 6 \; ; \; 2\pi \; ; \; \sqrt{51}$$