anonymous
  • anonymous
Given the equation 2 Square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution. x = 6, solution is not extraneous x = 6, solution is extraneous x = 11, solution is not extraneous x = 11, solution is extraneous
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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AkashdeepDeb
  • AkashdeepDeb
@meganemma1995 What exacrly do you mean by 2square root? :D
anonymous
  • anonymous
2√x-5=2
anonymous
  • anonymous
@AkashdeepDeb

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AkashdeepDeb
  • AkashdeepDeb
Okay so it is \[2 * \sqrt{x-5} = 2\] SQUARE BOTH SIDES 4(x-5) = 4 DIVIDE BOTH SIDES BY 4 x-5 = 1 ADD 5 TO BOTH SIDES x = 6 Understood? :)
anonymous
  • anonymous
would it be extraneous or not? @AkashdeepDeb
anonymous
  • anonymous
@AkashdeepDeb a extraneous is: a solution to an equation that does not fit the requirements of the original equation. and a non-extraneous equations is: an equation that does work
anonymous
  • anonymous
was is extraneous?
anonymous
  • anonymous
@xapproachesinfinity is this extraneous or not?
xapproachesinfinity
  • xapproachesinfinity
i said it is not did you read what i wrote i have given you the reason why
anonymous
  • anonymous
this is a different equation though, let me see if I can figure it out and then tell me if im right or wrong?
xapproachesinfinity
  • xapproachesinfinity
Oh i didn't pay attention to the question i thought it was the one you asked before
anonymous
  • anonymous
im going to say its not
anonymous
  • anonymous
@xapproachesinfinity
xapproachesinfinity
  • xapproachesinfinity
and why it is not?
anonymous
  • anonymous
because there was no 6 in the equation from the start? @xapproachesinfinity
xapproachesinfinity
  • xapproachesinfinity
the simpliest why to know if it is or not is to check your answer in the equation does work or not
xapproachesinfinity
  • xapproachesinfinity
no that's not the reason
anonymous
  • anonymous
I don't get it
xapproachesinfinity
  • xapproachesinfinity
do this to check \(\Large \tt\color{blueviolet}{2\sqrt{6-5}=2}\) we have to ask our selves is this true if it is true than the x=6 is not extraneous solution
xapproachesinfinity
  • xapproachesinfinity
it is a matter of checking if your answer satisfies the equation, yes?
xapproachesinfinity
  • xapproachesinfinity
So what do you think, actually phi has explained it in a better and sufficient way by now you should know how to do it^_^
xapproachesinfinity
  • xapproachesinfinity
\(\Large \tt\color{blueviolet}{2\sqrt{6-5}=2\sqrt{1}=2}\) so the solution works therefore it is not extraneous
xapproachesinfinity
  • xapproachesinfinity
you get it now?
anonymous
  • anonymous
in order to figure out whether something is extraneous or not extraneous, is to basically plug the variable back into the equation after you have already solved for the variable, right?

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