Question

Compare the irrational numbers: $3,\sqrt[3]{38},\sqrt{2},\sqrt[3]{3125},\sqrt{31},6$

• A. $6>\sqrt{31}>3>\sqrt[3]{38}>\sqrt{2}>\sqrt[3]{3125}$
• B. $6>\sqrt{31}>\sqrt[3]{3125}>3>\sqrt[3]{38}>\sqrt{2}$
• C. $6>3>\sqrt{31}>\sqrt[3]{38}>\sqrt{2}>\sqrt[3]{3125}$

Answer 1

The correct option is B $6>\sqrt{31}>\sqrt[3]{3125}>3>\sqrt[3]{38}>\sqrt{2}$

Let's simplify each term.

$\sqrt[3]{38}=\sqrt[3]{32^3}=2$

As, $3>2>\sqrt{2}\Rightarrow 3>\sqrt[3]{38}>\sqrt{2}$

$\therefore 3>\sqrt[3]{38}>\sqrt{2}$

Now, $\sqrt[3]{3125}=\sqrt[3]{35^3}=5$

From the perfect square method,
$\because 25<31<36\Rightarrow 5<\sqrt{31}<6$

Expressiong all the terms together we get,
$6>\sqrt{31}>\sqrt[3]{3125}>3>\sqrt[3]{38}>\sqrt{2}$