Question

PLEASE HELP NOT QUITE SURE ON THIS QUESTION Given the equation −4Square root of x minus 3 = 12, solve for x and identify if it is an extraneous solution.

Answer 1

this is the equation: \[\sqrt[-4]{x-3} = 12\]

Answer 2

\[\huge\rm -4\sqrt{x-3}=12\]
first move the -4 to the right side

Answer 3

it's not 4th root right ?
is it -4 (sqrt{x-3} ??

Answer 4

yes you have it right

Answer 5

@Nnesha @imqwerty

Answer 6

okay then first move the -4 to the right side how would do that ?
divide , multiply add , or subtract ?

Answer 7

you would add

Answer 8

@Nnesha

Answer 9

now it's -4 times sqrt{x-3}
what is opposite of multiplication ?

Answer 10

division

Answer 11

yes right so divide both side by -4

Answer 12

x-3=-2?

Answer 13

12/-4 = ??

Answer 14

-3

Answer 15

\[\sqrt{x-3}=-3\]
now take square both sides to cancel out the square root

Answer 16

so it would be 0

Answer 17

how ??

Answer 18

wait no -9

Answer 19

sorry

Answer 20

x-3=-9

Answer 21

\[\huge\rm (\sqrt{x-3})^2=(-3)^2\] when you take even power of NEGATIVE you will ALAWAYS get positive answer

Answer 22

oh okay

Answer 23

sp (-3)^2 =?
it's same as -3 times -3

Answer 24

positive 9

Answer 25

x-3= 9 solve for x

Answer 26

and then you would add 3 to get positive 12

Answer 27

and then would it be extraneous or not extraneous?

Answer 28

okay to check if it's extraneous or not
substitute x for 12 in the original equation

Answer 29

if both sides are equal then 12 isn't extraneous

Answer 30

both sides came out equal so its not extraneous

Answer 31

how did you get equal sides ?

Answer 32

honestly i don't know

Answer 33

\[-4\sqrt{12-3}=12\]
now solve

Answer 34

show me how did you do it ?

Answer 35

ok

Answer 36

im not sure how i did it but could you show me how to do it the right way please?

Answer 37

i just guessed because i wasn't sure how to do it, the 4 on the outside of the sqr root confuses me

Answer 38

okay so we replaced x with 12 right
now solve \[-4\sqrt{12-3}=12\]
12-3 and then take square root

Answer 39

seee i knew it
remember it's not 4 it's -4

Answer 40

hint: \[-1\cancel{=}1\]

Answer 41

i got \[\sqrt[16]{9} =12\]

Answer 42

that 16 isnt an exponent

Answer 43

hmm how did you get 16?
12-3 = ??

Answer 44

i did -4^2

Answer 45

don't take square \[-4\sqrt{12-3}\]
you should substitute x for 12 into the original equation

Answer 46

12-3= 9 right \[-4\sqrt{9}\] now solve

Answer 47

equals -12

Answer 48

yes so -12 =12 ?

Answer 49

sorry as you can tell i struggle with algebra

Answer 50

no so they are extraneous

Answer 51

the solution is not extraneous

Answer 52

solution is extraneous right ? @Nnesha

Answer 53

which one is it extraneous or not ?
like i said if you get equal sides THEN 12 is not extraneous

Answer 54

-12=12 both sides are equal ?

Answer 55

no they are not equal so it is extraneous

Answer 56

yes right

Answer 57

thanks so much for all your help that took awhile but thank you

Answer 58

my pleasure. :=)