Question

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

Answer 1

@meganemma1995 What exacrly do you mean by 2square root? :D

Answer 2

2√x-5=2

Answer 3

@AkashdeepDeb

Answer 4

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? :)

Answer 5

would it be extraneous or not? @AkashdeepDeb

Answer 6

@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

Answer 7

was is extraneous?

Answer 8

@xapproachesinfinity is this extraneous or not?

Answer 9

i said it is not
did you read what i wrote
i have given you the reason why

Answer 10

this is a different equation though, let me see if I can figure it out and then tell me if im right or wrong?

Answer 11

Oh i didn't pay attention to the question
i thought it was the one you asked before

Answer 12

im going to say its not

Answer 13

@xapproachesinfinity

Answer 14

and why it is not?

Answer 15

because there was no 6 in the equation from the start? @xapproachesinfinity

Answer 16

the simpliest why to know if it is or not
is to check your answer in the equation does work or not

Answer 17

no that's not the reason

Answer 18

I don't get it

Answer 19

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

Answer 20

it is a matter of checking if your answer satisfies the equation, yes?

Answer 21

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^_^

Answer 22

\(\Large \tt\color{blueviolet}{2\sqrt{6-5}=2\sqrt{1}=2}\) so the solution works
therefore it is not extraneous

Answer 23

you get it now?

Answer 24

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?