Question

Simplest rationalising factor of 5 root 128 is

Answer 1

Answer:

it is

$\sqrt{2}\\\sqrt{128} =\sqrt{2*2*2*2*2*2*2}\\\sqrt{2}$multiply this on both side to get rational number

so now

$\sqrt{128*2} = \sqrt{2*2*2*2*2*2*2*2}$

so

$\sqrt{256}=16^{2}\\ 5\sqrt{256}=5*16=80$

hope it helps

Answer 2

Answer:

The smallest rationalizing factor of  $\sqrt[5]{128}$  = $\sqrt[5]{2^3}$

Step-by-step explanation:

To find,

The simplest rationalizing factor of $\sqrt[5]{128}$

Solution:

Recall the concept:

The rationalizing factor is the number with which the irrational number to be multiplied to get a rational number

$a^mXa^n = a^{m+n}$

$a^\frac{1}{n}= \sqrt[n]{x}$

The prime factorization of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁷

$\sqrt[5]{128}$ = $\sqrt[5]{2^7}$

= $2^{\frac{7}{5} }$ ( by applying $a^\frac{1}{n}= \sqrt[n]{x}$)

= $2^{\frac{5+2}{5} }$

= $2^{\frac{5}{5} }$×$2^{\frac{2}{5} }$ ( by applying $a^m$×$a^n = a^{m+n}$)

= 2×$2^{\frac{2}{5} }$

Here the irrational part is $2^{\frac{2}{5} }$.

The smallest number to be multiplied to get a rational number is $2^{\frac{3}{5} }$ = $\sqrt[5]{2^3}$

∴ The smallest rationalizing factor of  $\sqrt[5]{128}$  = $\sqrt[5]{2^3}$

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