Question

The ascending order of $\sqrt{2},\sqrt[3]{34},\sqrt[4]{46}$

• A. $\sqrt{2},\sqrt[3]{34},\sqrt[4]{46}$
• B. $\sqrt{2},\sqrt[4]{46},\sqrt[3]{34}$
• C. $\sqrt[3]{34},\sqrt{2},\sqrt[4]{46}$
• D. $\sqrt[4]{46},\sqrt[3]{34},\sqrt{2}$

Answer 1

The correct option is B $\sqrt{2},\sqrt[4]{46},\sqrt[3]{34}$

The given surds can be rewritten as follows:

$\sqrt{2}={\left(2\right)}^{\frac{1}{2}}\sqrt[3]{34}={\left(4\right)}^{\frac{1}{3}}\sqrt[4]{46}={\left(6\right)}^{\frac{1}{4}}$

Now, since the LCM of the denominators of the powers is LCM$(2,3,4)=12$, therefore, we make the denominators same as shown below:

$\sqrt{2}={\left(2\right)}^{\frac{1}{2}}={\left(2\right)}^{\frac{1\times 6}{2\times 6}}={\left(2\right)}^{\frac{6}{12}}=\sqrt[12]{12{\left(2\right)}^6}=\sqrt[12]{1264}\sqrt[3]{34}={\left(4\right)}^{\frac{1}{3}}={\left(4\right)}^{\frac{1\times 4}{3\times 4}}={\left(4\right)}^{\frac{4}{12}}=\sqrt[12]{12{\left(4\right)}^4}=\sqrt[12]{12256}\sqrt[4]{46}={\left(6\right)}^{\frac{1}{4}}={\left(6\right)}^{\frac{1\times 3}{4\times 3}}={\left(6\right)}^{\frac{3}{12}}=\sqrt[12]{12{\left(6\right)}^3}=\sqrt[12]{12216}$

Since, $64<216<256$

Hence, the ascending order is $\sqrt{2}<\sqrt[4]{46}<\sqrt[3]{34}$