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This is one of our many quizzes on nth root functions. Click on this to access more such competitive quizzes on Number System.

nth root functions Quiz 2

mathematics

$  \sqrt[2]{{3x}} = 9 \\   \Rightarrow {\left( {\sqrt[2]{{3x}}} \right)^2} = {9^2} \\   \Rightarrow 3x = 81 \Rightarrow x = 27 \\  $

 The domain of $\sqrt[n]{x}$, where $n$ is an even natural number, is

 The even root of a negative number is not real.

For any negative value of $x$, the given function will become imaginary.

$  4\sqrt[4]{{x - 4}} = 20 \\   \Rightarrow \sqrt[4]{{x - 4}} = 5 \\   \Rightarrow {\left( {\sqrt[4]{{x - 4}}} \right)^4} = {5^4} \\   \Rightarrow x - 4 = 625 \Rightarrow x = 629 \\  $

 The implied domain of $y\left( x \right) = \sqrt {25 - {x^2}} $ is

 The square root of the negative number does not exist.

The square root of a negative number does not exist.  Thus, $ \\ 25 - {x^2} \geqslant 0 \\   \Rightarrow \left( {5 - x} \right)\left( {5 + x} \right) \geqslant 0 \\  $ The solution is $x \in \left[ { - 5,5} \right]$

 Which of these is among the ${5^{th}}$ roots of $2$?

 Using deMoivre’s Theorem to find the ${5^{th}}$ roots of $2$.

Let ${z^5} = 2$ then, ${z^5} = 2\left( {\cos 2k\pi  + i\sin 2k\pi } \right)\;\;,k = 0,1,2,3,4$ $ \\ $ Using deMoivre’s Theorem, $  z = {\left( {2\left( {\cos 2k\pi  + i\sin 2k\pi } \right)} \right)^{\dfrac{1}{5}}}\;\;,k = 0,1,2,3,4 \\   \Rightarrow z = {2^{\dfrac{1}{5}}}\left( {\cos \dfrac{{2k\pi }}{5} + i\sin \dfrac{{2k\pi }}{5}} \right)\;\;,k = 0,1,2,3,4 \\  $ Thus the ${5^{th}}$ roots of $2$ are, ${2^{\dfrac{1}{5}}}$, ${2^{\dfrac{1}{5}}}\left( {\cos \dfrac{{2\pi }}{5} + i\sin \dfrac{{2\pi }}{5}} \right)$, ${2^{\dfrac{1}{5}}}\left( {\cos \dfrac{{4\pi }}{5} + i\sin \dfrac{{4\pi }}{5}} \right)$, ${2^{\dfrac{1}{5}}}\left( {\cos \dfrac{{6\pi }}{5} + i\sin \dfrac{{6\pi }}{5}} \right)$ and ${2^{\dfrac{1}{5}}}\left( {\cos \dfrac{{8\pi }}{5} + i\sin \dfrac{{8\pi }}{5}} \right) \\  {2^{\dfrac{1}{5}}}\left( {\cos \dfrac{{6\pi }}{5} + i\sin \dfrac{{6\pi }}{5}} \right) = {2^{\dfrac{1}{5}}}\left( {\cos \left( {\pi  + \dfrac{\pi }{5}} \right) + i\sin \left( {\pi  + \dfrac{\pi }{5}} \right)} \right) \\   =  - {2^{\dfrac{1}{5}}}\left( {\cos \dfrac{\pi }{5} + i\sin \dfrac{\pi }{5}} \right) \\  $