Question

Compare the following pairs of surds. ${}^4\sqrt {64} ,{}^6\sqrt {128} $

Answer 1

Hint: Surds are defined as the values in square root i.e. within the radical $\sqrt {} $ which cannot be further simplified or broken down into whole numbers or integers. So, surds are irrational numbers. Some examples of surds are $\sqrt 2 ,\sqrt 3 ,\sqrt 5 $, etc., as we cannot simplify these values any further. On further simplification, they will be converted into decimal numbers.

Complete step-by-step answer:
We have given two surds ${}^4\sqrt {64} ,{}^6\sqrt {128} $ and we have to compare the pair of surds.
So we first need to expand the number within the radical $\sqrt {} $.
Initially, we take the surd ${}^4\sqrt {64} $ and perform the prime factorization of the number within the surd.
${}^4\sqrt {64} $
$ = {}^4\sqrt {2 \times 2 \times 2 \times 2 \times 2 \times 2} $
$ = {}^4\sqrt {{2^6}} $
We take four $2's$ and bring one $2$ outside the radical using the exponential formula ${}^n\sqrt {{a^m}} = {a^{\dfrac{m}{n}}}$.
$ = {}^4\sqrt {{2^4} \times {2^2}} $
$ = {\left( {{2^4}} \right)^{\dfrac{1}{4}}} \times {\left( {{2^2}} \right)^{\dfrac{1}{4}}}$
$ = 2 \times {2^{\dfrac{1}{2}}}$
$ = 2\sqrt 2 $
Here we manipulate the value to bring the number in the form of the second number. This will ensure that we will be able to compare the two values.
$ = 2{}^6\sqrt {{2^3}} $
$ = 2{}^6\sqrt 8 $
Now, we take the surd ${}^6\sqrt {128} $ and perform the prime factorization of the number within the surd.
${}^6\sqrt {128} $
$ = {}^6\sqrt {2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2} $
We take six $2's$ and bring one $2$ outside the radical using the exponential formula ${}^n\sqrt {{a^m}} = {a^{\dfrac{m}{n}}}$.

$ = {}^6\sqrt {{2^7}} $
$ = {}^6\sqrt {{2^6} \times 2} $
$ = {\left( {{2^6}} \right)^{\dfrac{1}{6}}} \times {\left( 2 \right)^{\dfrac{1}{6}}}$
$ = 2 \times {2^{\dfrac{1}{6}}}$
$ = 2{}^6\sqrt 2 $
We know $8 > 2$.
$\therefore \sqrt 8 > \sqrt 2 $
$ \Rightarrow {}^6\sqrt 8 > {}^6\sqrt 2 $
$ \Rightarrow 2{}^6\sqrt 8 > 2{}^6\sqrt 2 $
$ \Rightarrow {}^4\sqrt {64} > {}^6\sqrt {128} $
Thus, we observe that ${}^4\sqrt {64} $ is greater than ${}^6\sqrt {128} $.

Note: The power of both the surds must be equal while we compare the two surds. Else the comparison will be inappropriate and we will not get the correct result. One should be well aware of prime factorization while comparing surds. We need to write the numbers within the surds in exponential form so that the surds get simplified and it becomes easy to compare.