Question

One more for good measure? @jim_thompson5910
Find the exact value of the radical expression in simplest form.
the square root of s to the eighth power + the square root of 25 times s to the eighth power + 2 the square root of s to the eighth power − the square root of s to the fourth power

Answer 1

\[\sqrt{s^8} + \sqrt{25s^8}+2\sqrt{s^8}-\sqrt{s^4}\]

Answer 2

The things that look like 8's are s's

Answer 3

thanks

Answer 4

the first thing to do is to split up

\[\large \sqrt{25s^8}\]

into

\[\large \sqrt{25}*\sqrt{s^8}\]

Answer 5

what is the square root of 25?

Answer 6

Five

Answer 7

so

\[\large \sqrt{s^8} + \sqrt{25s^8}+2\sqrt{s^8}-\sqrt{s^4}\]

becomes

\[\large \sqrt{s^8} + 5\sqrt{s^8}+2\sqrt{s^8}-\sqrt{s^4}\]

Answer 8

now how do we simplify things like \(\large \sqrt{s^8} \) and \(\large \sqrt{s^4} \) ??

Answer 9

I forgot

Answer 10

what is the square root of x^2

Answer 11

x? I'm not really good with simplifying radical expressions, I had a mod get frustrated with me yesterday because I was terrible at it.

Answer 12

its ok, take it at your pace, not the mods

Answer 13

square roots undo squaring, so yes, \(\large \sqrt{x^2} = x\)

Answer 14

this is assuming that x is not negative

Answer 15

in general, the rule is this

\[\large \sqrt{x^n} = x^{n/2}\]

Answer 16

Okay

Answer 17

what that means is that

\[\large \sqrt{s^8} = s^{8/2} = s^4\]

and

\[\large \sqrt{s^4} = s^{4/2} = s^2\]

Answer 18

Would the answer be \[28s^4\sqrt{s}-s^2 ?\]

Answer 19

notice how

\[\large \sqrt{s^2} = s^{2/2} = s^1 = s\]

so the rule holds up to the fact that square rooting a square undoes it

Answer 20

so using those two facts, we can say

\[\large \sqrt{s^8} + 5\sqrt{s^8}+2\sqrt{s^8}-\sqrt{s^4}\]

turns into

\[\large s^4 + 5s^4 + 2s^4 - s^2\]

then you combine like terms to get

\[\large 8s^4 - s^2\]

Answer 21

so you were close in a way

Answer 22

Oopsies, alright. :o ^_^ Thanks again.

Answer 23

sure thing