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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.