Question

What value of x is an extraneous solution to

Answer 1

\[\sqrt{2x-8}=x-8\]

Answer 2

square and get
\[2x-8=x^2-16x+64\] then solve

Answer 3

you got this or no?

Answer 4

no is a fine answer just asking
i can walk you through it if you like

Answer 5

ye splease

Answer 6

ok
do you see how i got from
\[\sqrt{2x-8}=x-8\] to\[2x-8=x^2-16x+64\]?

Answer 7

yes

Answer 8

now set equal to zero so we have a quadratic equation to solve
subtract \(2x\) and add \(8\) to get
\[x^2-12x+72=0\]

Answer 9

ok

Answer 10

well it would be better if i did it right, should be
\[x^2-18x+72=0\]

Answer 11

ok

Answer 12

then do I find two number that multiply to be 72 and add to be -18?

Answer 13

yes

Answer 14

let me know when you find them
i am lousy at factoring but this should not be too hard

Answer 15

uuummm that would be -8 and -9?

Answer 16

no since \(-8-9=-17\)

Answer 17

close though
try one smaller and one bigger

Answer 18

-7and -10

Answer 19

no because that would be -17

Answer 20

-12 and -6

Answer 21

yea, and also because \(-7\times -10=-70\) and not \(-72\)

Answer 22

yes

Answer 23

by some miracle this factors as
\[(x-6)(x-12)=0\]

Answer 24

so
\[x=6\] or \[x=12\]
now one of those works, and one is "extraneous"

Answer 25

do I have to put in the number in to the actual equation?

Answer 26

yes

Answer 27

you can do it in your head

Answer 28

the answer is 6

Answer 29

yes, if by "answer' you mean the one that DOES NOT work
that is the extraneous one

Answer 30

yes

Answer 31

ok done
hope it was more or less clear

Answer 32

yes it was clear thank you

Answer 33

yw