anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@@peachpi @imqwerty @arindameducationusc @godgirl122 @heretohelpalways please help I'm confused
anonymous
  • anonymous
i did already but i got 2x-5=4
anonymous
  • anonymous
or would it be 4x-5=4?

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More answers

imqwerty
  • imqwerty
is this the question - \[\sqrt[2]{x-5}=2 \]??
anonymous
  • anonymous
yes
imqwerty
  • imqwerty
ok so when u square both the sides \[(\sqrt[2]{x+5})^2 =2^2\] what will be the RHS after squaring? :)
anonymous
  • anonymous
sorry , whats the RHS again?
imqwerty
  • imqwerty
RHS=right hand side :)
anonymous
  • anonymous
ok thanks it would be
imqwerty
  • imqwerty
right hand side of the equation
anonymous
  • anonymous
4
imqwerty
  • imqwerty
yes and what wuld be the Left hand side?
anonymous
  • anonymous
x+5
imqwerty
  • imqwerty
so we get x+5 =4
anonymous
  • anonymous
yes, now we solve for the variable and get -1 right?
anonymous
  • anonymous
@imqwerty
anonymous
  • anonymous
please help i don't know what to do after this step
imqwerty
  • imqwerty
yes u get x=-1 :)
anonymous
  • anonymous
@imqwerty wait don't go please help
imqwerty
  • imqwerty
ok
imqwerty
  • imqwerty
u need help in??
anonymous
  • anonymous
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
anonymous
  • anonymous
and none of those are -1 so i don't know what to do next
anonymous
  • anonymous
@imqwerty @peachpi
imqwerty
  • imqwerty
wait
imqwerty
  • imqwerty
we got x-5 = 1 add 5 to both sides u get x=6
anonymous
  • anonymous
ok thank you so much
anonymous
  • anonymous
how do i know if it is extraneous or not
imqwerty
  • imqwerty
and the solution is not extraneous cause when u put x=6 is a true solution :)
anonymous
  • anonymous
ok so if its a true solution its always not extraneous?
imqwerty
  • imqwerty
yes :)
anonymous
  • anonymous
ok, thank you for all of your help! it was much appreciated
imqwerty
  • imqwerty
no prblem :)

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