Question

Which one of the following numbers is a surd?
 $64, \sqrt {64}, \sqrt[3]{{64}}, \sqrt[4]{{64}}$

Answer 1

Hint: Here, we will proceed with the help of the prime factorisation method in order to solve the square root, cube root and a root of degree four given in the problem. If the finally obtained number appears to be irrational then that number is a surd.

Complete step-by-step answer:
We are given by four numbers i.e., $64, \sqrt {64}, \sqrt[3]{{64}}, \sqrt[4]{{64}}$
When we can’t simplify a number to remove a square root (or cube root or any higher degree root) then that number is called a surd. Surd are the numbers which consist of an irrational number.
Let us observe the first number which is 64. This number i.e., 64 can be represented as under.
As, $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^6}$
Clearly, this number 64 doesn’t consist of any irrational term like square root or cube root or any higher degree root. So, number 64 is not a surd.
Now, let us observe the second number which is $\sqrt {64} $. This number i.e., $\sqrt {64} $ can be represented as under.
As, $\sqrt {64} = \sqrt {2 \times 2 \times 2 \times 2 \times 2 \times 2} = \sqrt {{2^6}} = {\left( 2 \right)^{\dfrac{{6 \times 1}}{2}}} = {2^3} = 8$
Clearly, this number $\sqrt {64} $ doesn’t consist of any irrational term like square root or cube root or any higher degree root because this number is reduced into a rational number rather than an irrational number. So, the number $\sqrt {64} $ is not a surd.
Now, let us observe the third number which is $\sqrt[3]{{64}}$. This number i.e., $\sqrt[3]{{64}}$ can be represented as under.
As, $\sqrt[3]{{64}} = \sqrt[3]{{2 \times 2 \times 2 \times 2 \times 2 \times 2}} = \sqrt[3]{{{2^6}}} = {\left( 2 \right)^{\dfrac{{6 \times 1}}{3}}} = {2^2} = 4$
Clearly, this number $\sqrt[3]{{64}}$ doesn’t consist of any irrational term like square root or cube root or any higher degree root because this number is reduced into a rational number rather than an irrational number. So, the number $\sqrt[3]{{64}}$ is not a surd.
Now, let us observe the fourth number which is $\sqrt[4]{{64}}$. This number i.e., $\sqrt[4]{{64}}$ can be represented as under.
As, \[\sqrt[4]{{64}} = \sqrt[4]{{2 \times 2 \times 2 \times 2 \times 2 \times 2}} = \sqrt[4]{{{2^6}}} = {\left( 2 \right)^{\dfrac{{6 \times 1}}{4}}} = {2^{\dfrac{3}{2}}} = \sqrt {{2^3}} = 2\sqrt 2 \]
Clearly, this number $\sqrt[4]{{64}}$ consists of irrational term like square root or cube root or any higher degree root because this number is reduced into an irrational number (i.e., \[2\sqrt 2 \]) rather than a rational number. So, the number $\sqrt[4]{{64}}$ is a surd.
Therefore, out of the given four numbers i.e., $64, \sqrt {64}, \sqrt[3]{{64}}, \sqrt[4]{{64}}$, the number $\sqrt[4]{{64}}$ is a surd.

Note: In this particular problem, for the prime factorisation method we have represented the numbers given inside the square root, cube root and the root of degree four in terms of its prime factors. For a square root pair of two same prime factors can be taken outside, for cube root pair of three same prime factors can be taken outside and for the root of degree four pair of four same prime factors can be taken outside.