math question..extraneous solutions.?

Question

what is extraneous solutions. and for this question...

-z=(square root of) -z+6
which of these soliustions is extraneous z= -3
z=2

Answer 1

if you plug in the three for the z, you have

-(-3) which is positive three

3=sqrt.-(-3)+6 which is positive three again plus six, 9 the sqrt. of nine is three therefore three equals three.

but when you plug in two it comes out to

-2 = the square root of four which is 2, NOT negative two therefore the extraneous solution is z=2.

Answer 2

You solve the equation after number of processes and eventually get to several answers (in this case, -3 and 2). Although you arrived at these figures through valid procedures, sometimes, some of the answers are not valid when you plug it into the original formula.

For example,
x=-3
-(-3)=sqrt(-3)-z+6
You know you can not take sqrt of a negative number. (you can, but you end up in imaginary numbers) Thus, this equation is invalid.

When you plug in the number into the original equation and it makes no sense, (such as this case), it is called the extraneous result.

By the way, are you sure your result set and the formula is correct? Neither of them make sense when plugged into the original equation.

Answer 3

put z= -3 and z=2
into original eqts.
whichever doesn't work is extraneous

-z=(square root of) -z+6
-(-3)=sq rt( --3+6)
+3=sq rt(+9)
true

-z=(square root of) -z+6
-2=sq rt( -2+6)
-2=/=sq rt(4)2
false
sq rt(4)=+2 (positive value)

extraneous solutions don't solve original eqt.

Answer 4

Umm... I want to say it would be equal parts. 12.5 liters of each. But I would urge you to try this equation as an experiment and not on paper. You need a nearly empty and a half empty hand soap pump. And a snack, this could take a while. Best wishes. I would kill to be able to help you but my brain is frazzled just reading this stuff.