(Help with extraneous solutions) Find the solution of sqrX+3+4=6 and determine if it is an extraneous solution.

x = 1; not extraneous
x = 1; extraneous
x = 29; extraneous
x = 29; not extraneous

Question

(Help with extraneous solutions) Find the solution of sqrX+3+4=6 and determine if it is an extraneous solution. 

 	
	x = 1; not extraneous
	x = 1; extraneous
	x = 29; extraneous
	x = 29; not extraneous

anonymous · Answer

Find the solution of  5sqrx+7= -10 and determine if it is an extraneous solution. 

 	
	x = –3; extraneous
	x = –3; not extraneous
	x = 11; extraneous
	x = 11; not extraneous

rainbow_dashie · Answer

https://ca.answers.yahoo.com/question/index?qid=20111122084148AALLtIj

anonymous · Answer

I'm not completely sure that answers my question.

rainbow_dashie · Answer

http://openstudy.com/study#/updates/52f71139e4b0512c88e66e80

anonymous · Answer

I think that helps a bit I'm still confused about how to go about solving it sorry

rainbow_dashie · Answer

oh ok

rainbow_dashie · Answer

The instructions are precise

anonymous · Answer

No I get that but I mean I don't quite understand the extraneous part.

rainbow_dashie · Answer

Oh ok

anonymous · Answer

Do you know whether it's extraneous or not because I figured out that the answer is 29.

anonymous · Answer

hey @Angelwings22 ,
assuming the question is:
$$\sqrt{x+3+4}=6$$
then:
$$\sqrt{x+7}=6$$
square both size to get:
x+7=36
x=29

now,  extraneous solution is a solution that DOES NOT SATISFY the original equation.
so we have to plug 29 into the original equation:
$$\sqrt{29+3+4}=6$$
$$\sqrt{36}=6$$
indeed. so this is not  extraneous.

can you do your second question by yourself ?