We will discuss here about the meaning of \(\sqrt[n]{a}\).
The expression \(\sqrt[n]{a}\) means ‘nth rrot of a’. So, (\(\sqrt[n]{a}\))^n = a.
Also, (a^{1/a})^{n} = a^{ n × 1/n} = a^{1} = a.
So, \(\sqrt[n]{a}\) = a^{1/n}.
Examples:
1. \(\sqrt[3]{8}\) = 8^{1/3}
= (2^{3})^{1/3}
= 2^{3 × 1/3}
= 2^{1}
= 2.
2. \(\sqrt[4]{9}\) = 9^{1/4}
= (3^{2})¼
= 3^{2 × ¼}
= 3^{1/2}
= √3.
Note: 3^{1/2} = \(\sqrt[2]{3}\). But \(\sqrt[2]{3}\) is also written as √3.
Solved examples on nth Root of a:
Express each of the following in the simplest form without radicals:
(i) \(\sqrt[4]{5^{2}}\)
(ii) \(\sqrt[n]{x^{m}}\)
(iii) \(\sqrt[3]{64^{-4}}\)
Solution:
(i) \(\sqrt[4]{5^{2}}\) = (5^{2})^{1/4}
= 5^{2 × 1/4}
(ii) \(\sqrt[n]{x^{m}}\) = (x^{m})^{1/n}
= x^{m × 1/n}
= x^{m/n}.
(iii) \(\sqrt[3]{64^{-4}}\) = (64^{-4})^{1/3}
= 64^{-4 × 1/3}
= 64^{-4/3}
= (4^{3})^{-4/3}
= 4^{3(-4/3)}
= 4^{-4}
= \(\frac{1}{4 × 4 × 4 × 4}\)
= \(\frac{1}{256}\).
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