Question

@ParthKohli

Answer 1

I am getting only two correct answers ,wait I will post

Answer 2

\[\huge \left| x^2-2x-8 \right|+\left| x^2+x+8 \right|=3\left| x+2 \right|\]

Answer 3

So I cancelled mod(x-2)
and I get
\[\huge \left| x-4 \right|+\left| x-1 \right|=3\]

Answer 4

Sorry, I'm back.

Answer 5

By solving this , I get
answers as 1 and 4
but I don't get -2 which is also the answer

Answer 6

That's obviously because you cancelled \(|x+2|\) :P

Answer 7

yeah , I know but you can't do without that

Answer 8

I would instead take that common and then continue.

Answer 9

\[ \Large \left| x+2 \right|(\left| x-4 \right|\left| x-1 \right|)=3\left| x+2 \right|\]

Answer 10

Yeah, now take all of it on one side.

Answer 11

\[\Large \left| x+2 \right|(\left| x-4 \right|+x-1)-3\left| x+2 \right|=0\]

Answer 12

Now take |x+2| common.

Answer 13

oh I got it

Answer 14

And you should put a mod on |x-1| btw.

Answer 15

this equation tool sometimes is a pain , yeah

Answer 16

lol