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.

- anonymous

- schrodinger

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- anonymous

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

- Nnesha

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

- Nnesha

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

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## More answers

- anonymous

yes you have it right

- Nnesha

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

- anonymous

you would add

- anonymous

- Nnesha

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

- anonymous

division

- Nnesha

yes right so divide both side by -4

- anonymous

x-3=-2?

- Nnesha

12/-4 = ??

- anonymous

-3

- Nnesha

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

- anonymous

so it would be 0

- Nnesha

how ??

- anonymous

wait no -9

- anonymous

sorry

- anonymous

x-3=-9

- Nnesha

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

- anonymous

oh okay

- Nnesha

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

- anonymous

positive 9

- Nnesha

x-3= 9 solve for x

- anonymous

and then you would add 3 to get positive 12

- anonymous

and then would it be extraneous or not extraneous?

- Nnesha

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

- Nnesha

if both sides are equal then 12 isn't extraneous

- anonymous

both sides came out equal so its not extraneous

- Nnesha

how did you get equal sides ?

- anonymous

honestly i don't know

- Nnesha

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

- Nnesha

show me how did you do it ?

- anonymous

ok

- anonymous

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

- anonymous

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

- Nnesha

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

- Nnesha

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

- Nnesha

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

- anonymous

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

- anonymous

that 16 isnt an exponent

- Nnesha

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

- anonymous

i did -4^2

- Nnesha

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

- Nnesha

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

- anonymous

equals -12

- Nnesha

yes so -12 =12 ?

- anonymous

sorry as you can tell i struggle with algebra

- anonymous

no so they are extraneous

- anonymous

the solution is not extraneous

- anonymous

solution is extraneous right ? @Nnesha

- Nnesha

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

- Nnesha

-12=12 both sides are equal ?

- anonymous

no they are not equal so it is extraneous

- Nnesha

yes right

- anonymous

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

- Nnesha

my pleasure. :=)

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