Question

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Answer 1

well, sometimes when you solve an equation and end up with an "x" value, the x-value doesn't make the equation true when you plug it back in, because it isn't in the domain of the original function

in short: an extraneous solution is like a "fake" solution that you find that doesn't make the equation true when you plug it back in

Answer 2

so, looking at our problem, do you know how to solve for x?

Answer 3

can you subtract 1/(x+4)?

Answer 4

i like how i tag @vocaloid and3 other people come
xd

Answer 5

well, technically yes, but in this case we want to try a different method

this equation doesn't have any real (non-extraneous solutions) but we want to find the "fake" extraneous one

Answer 6

okay then idk

Answer 7

so, we multiply everything by (x+4) to get rid of the denominator

Answer 8

oohhhhh okay

Answer 9

that gives us

1 + (1/2)(x+4) = 1

can you try solving for x now?

Answer 10

First can you subtract the one?

Answer 11

yes, just treat it like a normal equation

Answer 12

how do you multiply 1/2 by x+4

Answer 13

you don't have to

after you subtract 1 from both sides, we have

(1/2)(x+4) = 0

now divide both sides by (1/2), what do you get?

Answer 14

x+4=0
lol sorry i knew that i don't know why i forgot
You get -4 right?

Answer 15

so its B?

Answer 16

nope, not B, hang with me for a sec ;)

x = -4 is called an "extraneous solution"

although we got x = -4 as an answer, when we plug x = -4 back into the original equation, we can see that the denominator of the first fraction becomes 0, x = -4 isn't a "real" solution

Answer 17

oh A

Answer 18

right-o

Answer 19

can u help me with a few more?

Answer 20

sure, post a new q and tag me

Answer 21

k