anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
this is the equation: \[\sqrt[-4]{x-3} = 12\]
Nnesha
  • Nnesha
\[\huge\rm -4\sqrt{x-3}=12\] first move the -4 to the right side
Nnesha
  • Nnesha
it's not 4th root right ? is it -4 (sqrt{x-3} ??

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anonymous
  • anonymous
yes you have it right
anonymous
  • anonymous
@Nnesha @imqwerty
Nnesha
  • Nnesha
okay then first move the -4 to the right side how would do that ? divide , multiply add , or subtract ?
anonymous
  • anonymous
you would add
anonymous
  • anonymous
@Nnesha
Nnesha
  • Nnesha
now it's -4 times sqrt{x-3} what is opposite of multiplication ?
anonymous
  • anonymous
division
Nnesha
  • Nnesha
yes right so divide both side by -4
anonymous
  • anonymous
x-3=-2?
Nnesha
  • Nnesha
12/-4 = ??
anonymous
  • anonymous
-3
Nnesha
  • Nnesha
\[\sqrt{x-3}=-3\] now take square both sides to cancel out the square root
anonymous
  • anonymous
so it would be 0
Nnesha
  • Nnesha
how ??
anonymous
  • anonymous
wait no -9
anonymous
  • anonymous
sorry
anonymous
  • anonymous
x-3=-9
Nnesha
  • Nnesha
\[\huge\rm (\sqrt{x-3})^2=(-3)^2\] when you take even power of NEGATIVE you will ALAWAYS get positive answer
anonymous
  • anonymous
oh okay
Nnesha
  • Nnesha
sp (-3)^2 =? it's same as -3 times -3
anonymous
  • anonymous
positive 9
Nnesha
  • Nnesha
x-3= 9 solve for x
anonymous
  • anonymous
and then you would add 3 to get positive 12
anonymous
  • anonymous
and then would it be extraneous or not extraneous?
Nnesha
  • Nnesha
okay to check if it's extraneous or not substitute x for 12 in the original equation
Nnesha
  • Nnesha
if both sides are equal then 12 isn't extraneous
anonymous
  • anonymous
both sides came out equal so its not extraneous
Nnesha
  • Nnesha
how did you get equal sides ?
anonymous
  • anonymous
honestly i don't know
Nnesha
  • Nnesha
\[-4\sqrt{12-3}=12\] now solve
Nnesha
  • Nnesha
show me how did you do it ?
anonymous
  • anonymous
ok
anonymous
  • anonymous
im not sure how i did it but could you show me how to do it the right way please?
anonymous
  • 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
  • Nnesha
okay so we replaced x with 12 right now solve \[-4\sqrt{12-3}=12\] 12-3 and then take square root
Nnesha
  • Nnesha
seee i knew it remember it's not 4 it's -4
Nnesha
  • Nnesha
hint: \[-1\cancel{=}1\]
anonymous
  • anonymous
i got \[\sqrt[16]{9} =12\]
anonymous
  • anonymous
that 16 isnt an exponent
Nnesha
  • Nnesha
hmm how did you get 16? 12-3 = ??
anonymous
  • anonymous
i did -4^2
Nnesha
  • Nnesha
don't take square \[-4\sqrt{12-3}\] you should substitute x for 12 into the original equation
Nnesha
  • Nnesha
12-3= 9 right \[-4\sqrt{9}\] now solve
anonymous
  • anonymous
equals -12
Nnesha
  • Nnesha
yes so -12 =12 ?
anonymous
  • anonymous
sorry as you can tell i struggle with algebra
anonymous
  • anonymous
no so they are extraneous
anonymous
  • anonymous
the solution is not extraneous
anonymous
  • anonymous
solution is extraneous right ? @Nnesha
Nnesha
  • Nnesha
which one is it extraneous or not ? like i said if you get equal sides THEN 12 is not extraneous
Nnesha
  • Nnesha
-12=12 both sides are equal ?
anonymous
  • anonymous
no they are not equal so it is extraneous
Nnesha
  • Nnesha
yes right
anonymous
  • anonymous
thanks so much for all your help that took awhile but thank you
Nnesha
  • Nnesha
my pleasure. :=)

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