Question

PLEASE HELP, IM LOST ON THIS QUESTION Given the equation 2Square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution.

Answer 1

@@peachpi @imqwerty @arindameducationusc @godgirl122 @heretohelpalways please help I'm confused

Answer 2

i did already but i got 2x-5=4

Answer 3

or would it be 4x-5=4?

Answer 4

is this the question - \[\sqrt[2]{x-5}=2 \]??

Answer 5

yes

Answer 6

ok so when u square both the sides \[(\sqrt[2]{x+5})^2 =2^2\]
what will be the RHS after squaring? :)

Answer 7

sorry , whats the RHS again?

Answer 8

RHS=right hand side :)

Answer 9

ok thanks it would be

Answer 10

right hand side of the equation

Answer 11

4

Answer 12

yes and what wuld be the Left hand side?

Answer 13

x+5

Answer 14

so we get x+5 =4

Answer 15

yes, now we solve for the variable and get -1 right?

Answer 16

@imqwerty

Answer 17

please help i don't know what to do after this step

Answer 18

yes u get x=-1 :)

Answer 19

@imqwerty wait don't go please help

Answer 20

ok

Answer 21

u need help in??

Answer 22

well my answer choices are these :
x= 6, solution is not extraneous
x = 6, solution is extraneous
x = 11, solution is not extraneous
x = 11, solution is extraneous

Answer 23

and none of those are -1 so i don't know what to do next

Answer 24

@imqwerty @peachpi

Answer 25

wait

Answer 26

we got x-5 = 1
add 5 to both sides u get x=6

Answer 27

ok thank you so much

Answer 28

how do i know if it is extraneous or not

Answer 29

and the solution is not extraneous cause when u put x=6 is a true solution :)

Answer 30

ok so if its a true solution its always not extraneous?

Answer 31

yes :)

Answer 32

ok, thank you for all of your help! it was much appreciated

Answer 33

no prblem :)