Question

`6/sqrt(2), sqrt(8), 4sqrt(2)` in ascending order is ______.

पर्याय

• `4sqrt(2), sqrt(8), 6/sqrt(2)`

• `sqrt(8), 6/sqrt(2), 4sqrt(2)`

• `6/sqrt(2), 4sqrt(2), sqrt(8)`

• `sqrt(8), 4sqrt(2), 6/sqrt(2)`

Answer 1

`6/sqrt(2), sqrt(8), 4sqrt(2)` in ascending order is `underlinebb(sqrt(8), 6/sqrt(2), 4sqrt(2))`.

Explanation:

Step 1: Simplify each expression

1. `6/sqrt(2)`

= `6/sqrt(2) xx sqrt(2)/sqrt(2)` 

= `(6sqrt(2))/2`

= `3sqrt(2)`

2. `sqrt(8)`

= `sqrt(4 xx 2)` 

= `2sqrt(2)`

3. Remains as is `4sqrt(2)`

Step 2: Compare coefficients of `sqrt(2)`

• `sqrt(8) = 2sqrt(2)`
• `6/sqrt(2) = 3sqrt(2)`
• `4sqrt(2)`

Clearly, `2sqrt(2) < 3sqrt(2) < 4sqrt(2)`

Step 3: Rewrite in original form

`sqrt(8) < 6/sqrt(2) < 4sqrt(2)`