Question

Help me solve dis?

Answer 1

4sqrtx^10

Answer 2

@YanaSidlinskiy @sammixboo

Answer 3

You gotta learn how to use \(LaTeX\) dude.

\(4 \sqrt{x^{10}}\)

Is that it?

Answer 4

I dunno how to use latex /.\
and no...i was uspposed to come up with a division radical...is this one?

Answer 5

Lol!! I was trying. Not that quick, thoughxD! Haha!!

Answer 6

yoou should use wolfram dudey

Answer 7

http://openstudy.com/study#/updates/52e12f56e4b0942cc9de719e

Answer 8

Learn \(LaTeX\) there.

Answer 9

I need to show my work @frizzank but tanks for trying lil guy :D

Answer 10

@undeadknight26 So it's not the equation that igreen showed?

Answer 11

nothats just a possibleone... I need help to find a non-extraneous division radical and solve it.

Answer 12

Can you post the actual question? *Exact* I wanna see it.

Answer 13

Mr. Jones is asked to write two types of radical equations for the SAT Prep Course. The first solution to the radical equation must be extraneous. The second solution to the radical equation must be non-extraneous. Write one equation where the solution is extraneous. Then write a second equation where the solution is non-extraneous. Using complete sentences, explain each step when solving to justify your examples.

Answer 14

I have so far:
A type of radical equations could be:
sqrt(6(x-15)) = sqrtx
sqrt(6(x-15))^2 = sqrtx^2
6x - 15 = x
-x -x
5x - 15 = 0
+ 15 +15
5x = 15
/5 /5
x = 3
Which is extraneous because if you plug 3 in for x the equation is false.
Another type of radicals would be:

Answer 15

Good! Exaclty what I needed:)

Answer 16

A radical equation with an extraneous solution is:
\(1 = \sqrt{x/3} - 2\)

Answer 17

Look. Idk what u did there. \(3=3 = \sqrt{x/3}\)

9 = x/3
x = 27

Check:
\(3=3 = \sqrt{27/3}\)
\(3=3 = \sqrt{9}\)
\(3 = 3\)
Makes sense? Sort of?

Answer 18

*tilts head* I think...I think so...

Answer 19

Which makes it a non-extraneous solution. What are you confused one? Btw, I started solving it out from the first equation I posted right above where I solved it. (Sorry it takes me 4ever!!!)

Answer 20

Anything like this..Will do.
\(x = \sqrt{something}\) that leads to "Something" less than zero.
Or.... x < 0

Answer 21

The answer to what "MUST" be extraneous is: \(\sqrt{x-4} = -3\)

Answer 22

I unda stand completely :D

Answer 23

Yay!!!!! Good!!!! Awesome!!! I was just about to say...."Hopefully u understand" Hahaha!!

Answer 24

Can you check it to make sure i did it right?

Answer 25

1 = sqrt(x/3) - 2
3 = 3 = sqrt(x/3)
3 = 3 = sqrt(27/3)
3 = 3 = sqrt9
3 = 3?

Answer 26

Yep!! Nailed it!!:)

Answer 27

Thanks for being awesome banana <4

Answer 28

Welcome!!:) Well, aren't I awesome always? The number increased to 4!! Yay!! <9

Answer 29

I got a 9?! omg ;-;

Answer 30

Hahha!!! Lol!! Ur special deep down in my heart!! Kk. I have to go now;)<3

Answer 31

Awe ur makin me blush :p