Question

Can someone please check this for me?
Radical Equations with Extraneous Solutions

Answer 1

@jim_thompson5910

Answer 2

@kohai

Answer 3

@mathmale

Answer 4

@TheSmartOne

Answer 5

@Whitemonsterbunny17

Answer 6

@zepdrix

Answer 7

In number 1, there is no square root in your steps...
kind of confusing :o

Answer 8

@zepdrix when you click the boxes, the square root symbol disappears for some reason

Answer 9

Oh and the boxes scroll :) woops...
Didn't see your other steps.

Answer 10

@quickstudent you made a mistake with your check on #2 (when you checked x = 5)

Answer 11

when you do the check, make sure you state which x value you're checking

so say something like

`checking x = 0`
`sqrt(5x) - x = 0`
`sqrt(5*0) - 0 = 0`
`sqrt(0) - 0 = 0`
`0 - 0 = 0`
`0 = 0 ... true`

`checking x = 5`
`sqrt(5x) - 5 = 0`
`sqrt(5*5) - 5 = 0`
`sqrt(25) - 5 = 0`
`5 - 5 = 0`
`0 = 0 ... true`

Answer 12

OK, I'll do that. Besides that, is everything correct?
@jim_thompson5910

Answer 13

yes other than that, everything looks great

Answer 14

Thanks :)

Answer 15

np