Question

Arrange the following surds in ascending order of their values from top to bottom.

• A. $\sqrt[3]{324}$
• B. $\sqrt{3}\times \sqrt{2}$
• C. $2\sqrt{27}$
• D. $\sqrt{72}$

Answer 1

To compare the surds, first let's convert them in to their simplest form.
$\sqrt[3]{324}=\sqrt[3]{38\times 3}=\left(\sqrt[3]{38}\right)\left(\sqrt[3]{33}\right)=2\sqrt[3]{33}$
$\sqrt{3}\times \sqrt{2}=\sqrt{6}$
$2\sqrt{27}=2\sqrt{9\times 3}=2\times 3\sqrt{3}=6\sqrt{3}$
$\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}$

$\text{Clearly,}\sqrt{6}<2\sqrt[3]{33}<6\sqrt{2}<6\sqrt{3}$
$\therefore$ The order will be,

• $\sqrt{3}\times \sqrt{2}$
• $\sqrt[3]{324}$
• $\sqrt{72}$
• $2\sqrt{27}$