Option A:
[tex]\sqrt{17}, \sqrt[3]{14}[/tex] are irrational numbers.
Solution:
Given numbers:
[tex]$\sqrt{17} , 0, \sqrt[3]{8} , \frac{22}{7} , \sqrt{81} , \sqrt[3]{14}[/tex]
To which of these are irrational numbers:
Irrational:
An irrational number cannot be written in the form [tex]\frac{a}{b}[/tex], where a and b are integers and b is non-zero.
(1) [tex]\sqrt{17}[/tex] cannot be written as [tex]\frac{a}{b}[/tex].
So, it is a irrational.
(2) [tex]\sqrt[3]{8} =\sqrt[3]{2^3}[/tex]
Cube and cube roots are get canceled.
[tex]\sqrt[3]{8} =2[/tex]
2 is an integer. So it is not a irrational number.
(3) [tex]\frac{22}{7}[/tex] is in the form [tex]\frac{a}{b}[/tex].
So it is not a irrational number.
(4)[tex]\sqrt{81}=\sqrt{9^2}[/tex]
Square and square root are get canceled.
[tex]\sqrt{81}=9[/tex]
9 is an integer. So it is not a irrational number.
(5) [tex]\sqrt[3]{14}[/tex]
14 cannot be written as [tex]\frac{a}{b}[/tex] form.
So, it is a irrational number.
Therefore [tex]\sqrt{17}, \sqrt[3]{14}[/tex] are irrational numbers.
Option A is the correct answer.
