An expression is of type sqrt if it is a radical and the exponent has a denominator of 2.

An expression is of type sqrt(domain) if it is of type radext(domain) and the exponent has a denominator of 2.

Note that a square root of a product or quotient is not of type sqrt because it evaluates to a product or quotient of square roots.

When used in the second form, it is necessary to enclose sqrt in forward (unevaluation) quotes to prevent the sqrt function from being invoked.

$\mathrm{type}\left(5^{\frac{1}{2}}\,'\mathrm{sqrt}'\right)$

$\mathrm{true}$

$\mathrm{type}\left(5^{\frac{1}{2}}\,'\mathrm{sqrt}'\left(\mathrm{integer}\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(y^{\frac{1}{2}}\,'\mathrm{sqrt}'\left(\mathrm{name}\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(5^{\frac{1}{4}}\,\mathrm{sqrt}\right)$

$\mathrm{false}$

$\mathrm{type}\left({\left(x+7\right)}^{\frac{3}{2}}\,'\mathrm{sqrt}'\left(\mathrm{integer}\right)\right)$

$\mathrm{false}$

$\mathrm{type}\left({\left(x+7\right)}^{\frac{3}{2}}\,'\mathrm{sqrt}'\left(\mathrm{polynom}\right)\right)$

$\mathrm{true}$