Question

Check whether ${\left(\sqrt{3}+2\right)}^2$ is rational number.

Answer 1

Let us assume the given number be rational and we will write the given number in $\frac{p}{q}$ form:

$\Rightarrow {(\sqrt{3}+2)}^2=\frac{p}{q}$

$\Rightarrow (\sqrt{3}{)}^2+2\times 2\times \sqrt{3}+4=\frac{p}{q}$

$\Rightarrow 7+4\sqrt{3}=\frac{p}{q}$

$\Rightarrow 4\sqrt{3}=\frac{p}{q}-7$

$\Rightarrow \sqrt{3}=\frac{p-7q}{q}÷4$

$\Rightarrow \sqrt{3}=\frac{p-7q}{4q}$.

We observe that the $LHS$ side is irrational and the $RHS$ side is rational, which is not possible.

That is, a contradiction to our assumption.

Hence our assumption that the number is rational is false.

$\Rightarrow$ The given number is irrational .