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.

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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
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i did already but i got 2x-5=4
or would it be 4x-5=4?

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

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