One approximate square root is a finite representation of a irrational number. In many cases, when working with square roots, an estimate with a few decimal places is enough for our calculations.
The calculator is an important tool in this process. Its display, which has limited space, indicates a good approximation for non-exact square roots. But it is also possible to find these estimates without the aid of a calculator, as we will see below.
Read too: Rooting — all about the inverse potentiation operation
Topics of this article
- 1 - Summary on approximate square root
- 2 - Video lesson on approximate square root
- 3 - How is the approximate square root calculated?
- 4 - Differences between approximate square root and exact square root
- 5 - Solved exercises on approximate square root
Approximate square root summary
An inexact square root is an irrational number.
We can find approximate values for non-exact square roots.
The accuracy of the approximation depends on the number of decimal places used.
The approximation can be done in different ways, including with the aid of the calculator.
Finding a y approximation to the square root of x means that y² is very close to x, but y² is not equal to x.
Video lesson on approximate square root
How do you calculate the approximate square root?
There are different ways to calculate the approximation of a square root. One of them is the calculator! For example, when we write \(\sqrt{2}\) on the calculator and click on =, the resulting number is an approximation. The same goes with \(\sqrt{3}\) It is \(\sqrt{5}\), which are also non-exact square roots, that is, they are irrational numbers.
Another way is to use exact roots close to the studied non-exact root. This lets you compare the decimal representations and find a range for the non-exact root. Thus, we can test some values until we find a good approximation.
It sounds difficult, but don't worry: it's a testing process. Let's look at some examples.
Examples
Find an approximation to two decimal places for \(\mathbf{\sqrt{5}}\).
realize that \(\sqrt{4}\) It is \(\sqrt{9}\) are the closest exact roots of \(\sqrt{5}\). Remember that the larger the radicand, the larger the square root value. Thus, we can conclude that
\(\sqrt{4}
\(2
I.e, \(\sqrt5\) is a number between 2 and 3.
Now is the time for testing: we choose some values between 2 and 3 and check if each squared number approaches 5. (Remember that \(\sqrt5=a\) if \(a^2=5\)).
For the sake of simplicity, let's start with numbers with one decimal place:
\(2,1^2=4,41\)
\(2,2^2=4,84\)
\(2,3^2=5,29\)
Note that we don't even need to continue parsing numbers to one decimal place: the number we're looking for is between 2.2 and 2.3.
\(2,2
Now, as we are looking for an approximation with two decimal places, let's proceed with the tests:
\(2,21^2=4,8841\)
\(2,22^2=4,9284\)
\(2,23^2=4,9729\)
\(2,24^2=5,0176\)
Again, we can stop the analysis. The number you are looking for is between 2.23 and 2.24.
\(2.23
But and now? Which of these values with two decimal places do we choose as an approximation of \(\sqrt5\)? Both are good options, but note that the best is the one whose square is closest to 5:
\(5–2,23^2=5-4,9729=0,0271\)
\(2,24^2-5=5,0176-5=0,0176\)
I.e, \(2,24^2 \) is closer to 5 than \(2,23^2\).
Thus, the best approximation to two decimal places for \(\sqrt5\) é 2,24. We write that \(\sqrt5≈2.24\).
Do not stop now... There's more after the publicity ;)
Find an approximation to two decimal places for \(\mathbf{\sqrt{20}}\).
We could start in the same way as in the previous example, that is, look for exact roots whose radicands are close to 20, but note that it is possible to decrease the value of the radicand and facilitate the accounts:
\(\sqrt{20}=\sqrt{4·5}=\sqrt4·\sqrt5=2\sqrt5\)
Note that we performed the decomposition of the radicand 20 and used a rooting property.
Now how \(\sqrt20=2\sqrt5\), we can use the approximation with two decimal places to \(\sqrt5\) from the previous example:
\(\sqrt{20} ≈2.2,24 \)
\(\sqrt{20} ≈4.48\)
Observation: As we use an approximate number (\(\sqrt5≈2.24\)), the value 4.48 may not be the best approximation with two decimal places for \(\sqrt{20}\).
Read too: How to calculate the cube root of a number?
Differences between approximate square root and exact square root
An exact square root is a rational number. realize that \(\sqrt9\),\(\sqrt{0,16}\) It is \(\sqrt{121}\) are examples of exact square roots, as \(\sqrt{9}=3\), \(\sqrt{0,16}=0,4\) It is \(\sqrt{121}=11\). Furthermore, when we apply the inverse operation (that is, the potentiation with exponent 2), we get the radicand. In the previous examples, we have \(3^2=9\), \(0,4^2=0,16\) It is \(11^2=121\).
An inexact square root is an irrational number (that is, a number with infinite non-repeating decimal places). Thus, we use approximations in its decimal representation. realize that \(\sqrt2\), \(\sqrt3\) It is \(\sqrt6\) are examples of non-exact roots, because \(\sqrt2≈1.4142135\), \(\sqrt3≈1.7320508\) It is \(\sqrt6≈2.44949\). Furthermore, when we apply the inverse operation (that is, potentiation with exponent 2), we obtain a value close to the radicand, but not equal. In the previous examples, we have \(1,4142135^2=1,999999824\), \(1,7320508^2=2,999999974\) It is \(2,44949^2=6,00000126\).
Solved exercises on approximate square root
question 1
Arrange the following numbers in ascending order: \(13,\sqrt{150},\sqrt{144},14\).
Resolution
realize that \(\sqrt{150}\) is a non-exact square root and \(\sqrt{144}\) is exact (\(\sqrt{144}=12\)). Thus, we only need to identify the position of \(\sqrt{150}\).
note that \(13=\sqrt{169}\). Considering that the greater the radicand, the greater the value of the square root, we have that
\(\sqrt{144} < \sqrt{150} < \sqrt{169}\)
Therefore, arranging the numbers in ascending order, we have
\(\sqrt{144} < \sqrt{150} < 13 < 14\)
question 2
Among the following alternatives, which is the best approximation with one decimal place for the number \(\sqrt{54}\)?
a) 6.8
b) 7.1
c) 7.3
d) 7.8
e) 8.1
Resolution
Alternative C
note that \(\sqrt{49}\) It is \(\sqrt{64}\) are the closest exact square roots of \(\sqrt{54}\). As \(\sqrt{49}=7\) It is \(\sqrt{64}=8\), We have to
\(7
Let's see some possibilities of approximation with one decimal place for \(\sqrt{54}\):
\(7,1^2=50,41\)
\(7,2^2=51,84\)
\(7,3^2=53,29\)
\(7,4^2=54,76\)
Note that it is not necessary to continue with the tests. Also, among the alternatives, 7.3 is the best approximation to one decimal place for \(\sqrt{54}\).
By Maria Luiza Alves Rizzo
Math teacher
Click to check how the calculation of non-exact roots can be done by decomposing the radicand into prime factors!
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