In comparison of surds we will discuss about the comparison of equiradical surds and comparison of nonequiradical surds.
I. Comparison of equiradical surds:
In case of equiradical surds (i.e., surds of the same order) \(\sqrt[n]{a}\) and \(\sqrt[n]{b}\), we have \(\sqrt[n]{a}\) > \(\sqrt[n]{b}\) when x > y.
For example,
(i) √5 > √3, since 5 > 3
(ii) ∛21 < ∛28, since 21 < 28.
(iii) ∜10 > ∜6, since 10 > 6.
II. Comparison of nonequiradical surds:
In case of comparison between two or more nonequiradical surds (i.e., surds of different orders) we express them to surds of the same order (i.e., equiradical surds). Thus, to compare between ∛7 and ∜5 we express them to surds of the same order as follows:
Clearly, the orders of the given surds are 3 and 4 respectively and LCM Of 3 and 4 is 12.
Therefore, ∛7 = 7\(^{1/3}\) = 7\(^{4/12/}\) = \(\sqrt[12]{7^{4}}\) = \(\sqrt[12]{2401}\) and
∜5 = 5\(^{1/4}\) = 5\(^{3/12}\) = \(\sqrt[12]{5^{3}}\) = \(\sqrt[12]{125}\)
Clearly, we see that 2401 > 125
Therefore, ∛7 > ∜5.
Example of comparison of surds:
Convert each of the following surds into equiradical surds of the lowest order and then arrange them in ascending order.
∛2, ∜3 and \(\sqrt[12]{4}\)
Solution:
∛2, ∜3 and \(\sqrt[12]{4}\)
We see that the orders of the given surds are 3, 4 and 12 respectively.
Now we need to find the lowest common multiple of 3, 4 and 12.
The lowest common multiple of 3, 4 and 12 = 12
Therefore, the given surds are expressed as equiradical surds of the lowest order (i.e. 12th order) as follows:
∛2 = 2\(^{1/3}\) = 2\(^{4/12}\) = \(\sqrt[12]{2^{4}}\) = \(\sqrt[12]{16}\)
∜3 = 3\(^{1/4}\) = 3\(^{3/12}\) = \(\sqrt[12]{3^{3}}\) = \(\sqrt[12]{27}\)
\(\sqrt[12]{4}\) = 4\(^{1/12}\) = \(\sqrt[12]{4^{1}}\) = \(\sqrt[12]{4}\)
Therefore, equiradical surds of the lowest order ∛2, ∜3 and \(\sqrt[12]{4}\) are \(\sqrt[12]{16}\), \(\sqrt[12]{27}\) and \(\sqrt[12]{4}\) respectively.
Clearly, 4 < 16 < 27; hence the required ascending order of the given surds is:
\(\sqrt[12]{4}\), ∛2, ∜3
11 and 12 Grade Math
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