Write in descending order- 2 sqrt-4-6- and 3 sqrt-4-2-

Question

Write in descending order- 2 sqrt-4-6-  and  3 sqrt-4-2-

Toppr Admin · Accepted Answer

- 2 -sqrt-4-6-sqrt-4-2-times2-times2-times2 -times 6-sqrt-4-96- - 3 -sqrt-4-2-sqrt-4-3-times3-times3-times3 -times 2-sqrt-4-162- -We know that - 162-gt-96 -160- -Rightarrow -sqrt-4-162-gt-sqrt-4-96- - -Rightarrow 3 -sqrt-4-2-gt-2 -sqrt-4-6- -

Write in descending order:
$$ 2 \sqrt[4]{6} $$ and $$ 3 \sqrt[4]{2} $$

$$ 2 \sqrt[4]{6}=\sqrt[4]{2\times2\times2\times2 \times 6}=\sqrt[4]{96} $$
$$ 3 \sqrt[4]{2}=\sqrt[4]{3\times3\times3\times3 \times 2}=\sqrt[4]{162} $$
We know that $$ 162>96 $$
$$ \Rightarrow \sqrt[4]{162}>\sqrt[4]{96} $$
$$ \Rightarrow 3 \sqrt[4]{2}>2 \sqrt[4]{6} $$

Write in descending order:$$ 2 \sqrt[4]{6} $$ and $$ 3 \sqrt[4]{2} $$

$$ 2 \sqrt[4]{6}=\sqrt[4]{2\times2\times2\times2 \times 6}=\sqrt[4]{96} $$$$ 3 \sqrt[4]{2}=\sqrt[4]{3\times3\times3\times3 \times 2}=\sqrt[4]{162} $$We know that $$ 162>96 $$ $$ \Rightarrow \sqrt[4]{162}>\sqrt[4]{96} $$$$ \Rightarrow 3 \sqrt[4]{2}>2 \sqrt[4]{6} $$

Write in descending order:
$$ 2 \sqrt[4]{6} $$ and $$ 3 \sqrt[4]{2} $$

$$ 2 \sqrt[4]{6}=\sqrt[4]{2\times2\times2\times2 \times 6}=\sqrt[4]{96} $$
$$ 3 \sqrt[4]{2}=\sqrt[4]{3\times3\times3\times3 \times 2}=\sqrt[4]{162} $$
We know that $$ 162>96 $$
$$ \Rightarrow \sqrt[4]{162}>\sqrt[4]{96} $$
$$ \Rightarrow 3 \sqrt[4]{2}>2 \sqrt[4]{6} $$