Question

How do I solve this... Algebra two help? Medal?

Answer 1

First you need to get the expression with x in it out from under the square root sign, how would you do that?

Answer 2

square both sides?

Answer 3

Yes, to save computation divide both sides by- 4 first to get
\[\sqrt{x-3}=-3\]

Answer 4

ohhh okay so the answer would be x=12?

Answer 5

square both sides to get
x-3=9
x=12 is that an extraneous solution?

Answer 6

yes?

Answer 7

Well, 12 results in
\[-4\sqrt{12-3}=-4+/-3 when the \square \root value is +3 the solution of x=12 does \not work so \it is extraneous \in that case, but \it does work for -3\]

Answer 8

That didn't work sorry, for the -3 value of the square root the x=12 solution works, for the +3 it is extraneous.

Answer 9

So 12 is not extraneous??

Answer 10

It is when it leads to a +3 value, but it isn't when it leads to a -3 value

Answer 11

So in this case it is not! Thanks so much for all your help!!