Question

Solve. Check for extraneous solutions. Did I do this problem correctly?

1. (x+5)^1/2 - (5-2x)^1/4 = 0
((x+5)^1/2)^2 - ((5-2x)^1/4)^4 = 0
(x+5) - (5-2x) = 0
x+5 = 5 +2x
5 = 5+x
x = 0

I'm not sure if I did my math correctly.

Answer 1

nope

Answer 2

Oop I thought so. It didn't seem right. Do you mind explaining?

Answer 3

you cant multiply powers like that

Answer 4

hold on
i will tell

Answer 5

Thank you (:

Answer 6

\[(x+5)^{1/2} = (5-2x)^{1/4}\]

Answer 7

see i tok 5-2x term on right side

Answer 8

took*

Answer 9

now

Answer 10

multiply the powers by 4
\[(x+5)^{4/2}=(5-2x)^{4/4}\]

Answer 11

you get
\[(x+5)^2 = 5-2x\]
solve the rest
tell me the answer

Answer 12

understood?

Answer 13

Oh yes I see! Thank you very much! (:

Answer 14

so what are the answers?

Answer 15

I'm working on it. So far I have
x^2 +10+25=5-2x

Answer 16

okay
let me know when you are done

Answer 17

Okay I am done. I did:
x^2 + 12x + 20 = 0
(x+10)(x+2) = 0
x= -10 x=-2 BUT -10 is an extraneous solution because if you substitute it for x, it won't equal to 0.

Answer 18

why wouldn't it?
\[-10^2 +12(-10)+20 =100-120+20 =120-120=0\]

Answer 19

right?

Answer 20

I substituted it back to the original problem (x+5)^1/2 - (5-2x)^1/4 = 0 and the calculator just said domain error.

Answer 21

LOL
why to put it in original equation when you have simplified 1
-10 and -2 are your values

Answer 22

Yes that is correct. My paper just asked me to check for extraneous solutions and that's what I've been doing for my other problems. Thank you for the big help though (: I was so lost :o

Answer 23

in original equation -10+5 will give you -5
so sqrt of -5 is imaginary number
that's why it gives domain error

Answer 24

haha
you are welcome x)