In this explainer, we will learn how to approximate the square or cube root of a number and then use it to solve real-world problems.

We can directly calculate the square root of a perfect square and the cube root of a perfect cube, but we cannot directly evaluate the square root of a nonperfect square or the cube root of a nonperfect cube. The reason for this is that these numbers are irrational and so will not have a representation as a fraction nor a finite or repeating decimal expansion. Much like , these numbers have an infinite nonrepeating decimal expansion, so the best we can do is approximate the numbers to a certain number of digits.

To demonstrate this, let’s try to find an approximate value of . We know that , so we are looking for a number whose square is 13. We can try squaring some values.

First, let’s try 3. We see that

This is too low, so we will need to try a larger number.

Second, let’s try 4. We see that

This is too high, so we will need to try a smaller number.

Third, let’s try 3.5. We calculate that

12.25 is less than 13, so this is too small. Let’s now try 3.7. We calculate

This is too high, so we will try 3.6. We have

We can see that 3.6 is smaller than and that 3.7 is greater than . So, lies between 3.6 and 3.7.

There are some things worth noting about this method.

Firstly, this method works since using a larger argument in the square or cube root function will always increase the output. We can visualize this by considering a square or a cube and considering what happens to the side length when we increase the area or volume; the side length must increase.

In particular, we can note that a square with a side length of 3 will have an area of , a square with a side length of will have an area of , and a square with a side length of 4 will have an area of

This allows us to visualize why lies between 3 and 4.

Secondly, this method allows us to approximate these radical values to any number of decimal places, provided we can accurately calculate the squares or cubes of the approximation.

Thirdly, we often contain this information in a table to help keep track of approximations and their sizes relative to that of the radical. For example, we can keep our above approximations in the following table.

Comparison to 13 | ||
---|---|---|

3 | 9 | |

4 | 16 | |

3.5 | 12.25 | |

3.7 | 13.69 | |

3.6 | 12.96 |

Let’s now see an example of applying this method to determine the nearest integer upper and lower bounds of a radical.

### Example 1: Finding Two Consecutive Integers That a Radical Number Lies Between

Sameh is trying to find which two whole numbers lie on either side of . He decides that it will be helpful to use what he has been taught about square numbers.

What is the greatest square number below 51?

What is the smallest square number above 51?

Hence, determine the two consecutive whole numbers that lies between.

### Answer

**Part 1**

We start by listing the square numbers until they go above 51; these are given by 1, 4, 9, 16, 25, 36, 49, and 64. We can see that the largest of these values that is below 51 is 49.

Hence, the greatest square number below 51 is 49.

**Part 2**

We can also see from this list that the smallest square number above 51 is 64.

**Part 3**

In parts 1 and 2, we showed that . Since and , we must have that .

Hence, lies between 7 and 8.

In our next example, we will complete a table of values to help determine the approximation for that is accurate to two decimal places.

### Example 2: Estimating a Radical Number to Two Decimal Places Using a Table

Sameh wants to find an approximation for accurate to two decimal places. He sets up the equation and then starts using trial and improvement to find the solution, as seen in the table.

Big/Small | ||
---|---|---|

1 | small | |

2 | big | |

1.5 | big | |

1.4 | small | |

1.41 | ||

1.42 | ||

1.415 |

- Determine the value of , giving your answer accurate to three decimal places.
- Determine the value of , giving your answer accurate to three decimal places.
- Determine the value of , giving your answer accurate to three decimal places.
- Hence, what is the value of accurate to two decimal places?

### Answer

**Part 1**

We calculate this value directly; we have

To three decimal places, that is 1.988.

It is worth noting that this is smaller than 2.

**Part 2**

We can also calculate this value directly; we have

To three decimal places, that is 2.016.

It is worth noting that this value is bigger than 2.

**Part 3**

We calculate that

To three decimal places, that is 2.002.

Once again, it is worth noting that this is bigger than 2.

**Part 4**

Using parts 2 and 3, we have shown that . Therefore, . Since 1.415 is the smallest number possible that rounds up to give 1.42 to two decimal places and is less than 1.415, this means must round down to give 1.41 to two decimal places. Thus, to two decimal places.

Finding a decimal approximation is not the only way to represent a number. We can also find its position on a number line. This then gives us the question of how to find the points representing radicals on a number line. This is often called compass-and-straightedge construction, since we are allowed to use these tools to construct the points.

Although it is possible to determine the points representing cube roots on a number line using a compass and straightedge, it is beyond the scope of this explainer to show this method. Instead, we will focus entirely on square roots.

We will start with a number line with points marked for 0 and 1.

Let’s say we want to determine the point on the number line representing . We can do this by recalling that the Pythagorean theorem tells us that if the legs of a right triangle have lengths and , then , where is the length of the hypotenuse.

We can take the square roots of both sides of this equation to see that . Thus, if and , we have . Therefore, a right triangle with both legs of length 1 will have a hypotenuse of length . We can construct such a right triangle by adding a vertical line at 0 with a point at 1. It is worth noting that we can find the point at 1 using our compass with its point at 0 and the pencil at 1; then, the intersection of the circle with the vertical line is at 1. This gives us the following.

We can then construct a right triangle using the diagonal of the unit square as follows.

We know that the length of the hypotenuse is due to the Pythagorean theorem. We can now draw a semicircle of radius centered at 0 by putting our compass point at 0 and the pencil at the end of the line segment of length . Every point on this semicircle is a distance of from 0, so the points of intersection on the number line must be and . We label the point of intersection on the positive side as , as shown.

We can continue this process to find points representing other square roots. For example, if we wanted to find the point representing on the number line, we can note that . So, is the length of a hypotenuse of a right triangle with legs of lengths and 1. We can find such a right triangle using the diagonal of a rectangle of length and width 1, as shown.

We can then draw a semicircle of radius centered at 0 to find points of distance from 0. In particular, the two points of intersection between the number line and the semicircle will be the points representing and ; we label the point on the positive side , as shown.

Let’s now see an example of applying this method to determine the point representing on a number line.

### Example 3: Identifying the Exact Position of a Radical Expression on a Number Line

Which of the following is the correct construction of on a number line using a compass and a straightedge?

### Answer

We could answer this by eliminating choices; however, to fully understand the process, we will construct the point representing on the number line from first principles.

First, we need to determine a geometric relationship that involves . We can do this by noting that . Then, the Pythagorean theorem tells us that a right triangle with legs of lengths 2 and 3 will have a hypotenuse of length .

We want to construct a right triangle with legs of lengths 2 and 3. One way of doing this is to use a compass to mark points at a distance of 2 and 3 from 0 on a set of perpendicular axes. We can do this by constructing perpendicular lines, putting the point on 0 and the pencil on 1 to get a length of 1, and then successively marking points. We can then draw a rectangle using these points, as shown

The Pythagorean theorem then tells us that the diagonal of this rectangle has a length of .

We can then put the point of our compass at 0 and the pencil at the other end of this line segment of length to sketch a semicircle of points at a distance of from 0. The points of intersection between the semicircle and the horizontal number line will be and . We mark the point in the positive direction as , as shown.

This is the same figure as the answer given in choice B.

In our next example, we will find an approximate solution to an equation using the approximation of radicals.

### Example 4: Solving a Two-Step Equation and Approximating the Solution

Solve the equation , then approximate your answer to one decimal place.

### Answer

We start by noting that the left-hand side of the equation is the cube of a linear expression, so we can simplify the equation by taking cube roots of both sides; this gives

We can then solve for by adding 2 to both sides of the equation to get

We can then find an approximate value for by approximating to one decimal place and then adding 2. We note that , so . We can then check the cubes of other numbers to see which is closest to 6. We have

Thus,

To help us determine if we would round up or down, let’s consider :

We see that this is larger that 6, so

Hence, we would round the decimal expansion of down to get to one decimal place.

Therefore,

In our final example, we will apply this method of approximating radicals to a real-world example.

### Example 5: Estimating the Length of a Fence given the Area It Encloses

Fady wants to buy fencing to enclose a square
garden with an area of
380 ft^{2}. How much fencing does he need to buy? Round your answer to the nearest tenth.

### Answer

Let’s start by sketching the information given. The garden is square and has an area of
380 ft^{2}. If we say that the side length of the garden is
ft, we get the following.

We know that the area of a square is the square of its side length, so we must have

We can solve this equation for by noting that is nonnegative, since it is a length, and then taking the square roots of both sides of the equation to get

Before we approximate the value of , it is worth noting that we are asked to find the amount of fencing Fady needs to buy rounded to the nearest tenth. This is given by the perimeter of the garden, which is . Therefore, we are not rounding to the nearest tenth; we are rounding to the nearest tenth. We will instead round to two decimal places of accuracy to reduce inaccuracy when rounding.

We want to approximate to the nearest hundredth. To do this, we first note that and . So,

Hence,

We can check the squares of numbers between 19 and 20 to determine which is closest to 380. We have

Thus,

To determine if the decimal expansion of should be rounded up or down to two decimal places, we need to compare to 19.495. We calculate which is greater than 380. Therefore,

So, we must round the expansion down, which gives to two decimal places.

We then have

We then round this value to the nearest tenth to get 78.0 ft.

Let’s finish by recapping some of the important points from this explainer.

### Key Points

- We can calculate the square or cube root of a number to any number of decimal places by considering numbers whose squares are above and below .
- We can use the Pythagorean theorem and a compass and straightedge to find points representing radicals on a number line.
- We can find decimal approximations of a radical by comparing its size to those of the decimals. In general, we can take the halfway point of two approximations to improve our approximation. This also allows us to determine the approximation correct to any number of decimal places by approximating the radical with a decimal of one extra decimal place of accuracy so that we can determine if the value should be rounded up or down.