Given:
$\sqrt{3}$ is irrational.
To Prove:
$5+2\sqrt{3}$ is irrational.
Proof:
Suppose that $5+2\sqrt{3}$ is a rational number. Then, it can be written in $\large{\frac{p}{q}}$ form where integers $p$ and $q$ have no common prime factor.
$\hspace{1.75 cm}5+2\sqrt{3}=\large\frac{p}{q}$
$\text{Where, HCF}(p,q)=1\text{ and }q\neq 0.$
$\hspace{1.3 cm}2\sqrt{3} = \large{\frac{p}{q}}-5$
$\hspace{1.3 cm}2\sqrt{3} = \large{\frac{p-5q}{q}}$
$\hspace{1.3 cm}\sqrt{3} = \large{\frac{p-5q}{2q}}$
It shows that $\sqrt{3}$ can be written in $\large{\frac{p}{q}}$ form which cannot be possible because $\sqrt{3}$ is irrational. So, our assumption is wrong. Hence $4+2\sqrt{3}$ is irrational.