Examine whether the following numbers are rational or irrational:$ (\sqrt{2}+\sqrt{3})^{2} $


Given:

\( (\sqrt{2}+\sqrt{3})^{2} \)

To do:

We have to classify the given number as rational or irrational.

Solution:  

A rational number can be expressed in either terminating decimal or non-terminating recurring decimals and an irrational number is expressed in non-terminating non-recurring decimals.

Therefore,

$(\sqrt{2}+\sqrt{3})^{2}=(\sqrt{2})^2+(\sqrt{3})^2+2\times\sqrt{2}\times\sqrt{3}$

$=2+3+2\sqrt{2\times3}$

$=5+2\sqrt{6}$

$\sqrt{6}=2.4494897............$

The decimal expansion of $\sqrt{6}$ is non-terminating and non-recurring.

The sum of a rational number and an irrational number is an irrational number.

Therefore, \( (\sqrt{2}+\sqrt{3})^{2} \) is an irrational number.   

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Updated on: 10-Oct-2022

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