Question

Help ugh
Solve for x given (x+4)^4 - 5(x+4)^2+6=0

Answer 1

May I help?

Answer 2

PLEASE DO ! ! !

Answer 3

Thank you.

Answer 4

Hint: put \[y=(x+4)\] and tell me what you got?

Answer 5

(y^4) - 5(y^2) + 6 = 0

Answer 6

Great!
\[y^4-5y^2+6=0\]
If you can, it'b better to use the equation box to type in your expressions.
Now can you factor this equation into two factors firstly?

Answer 7

Need assistance?

Answer 8

would it be \[(y ^{2} - 2) (y ^{2}\pm 3)\]

Answer 9

why \[\pm \]?

Answer 10

because for the previous equation i got
\[y ^{2} (y ^{2}-3) -2(y ^{2}+3)\]

Answer 11

Man, it is just like that!
\[(y^2-2)(y^2-3)\]
and make a check by expand it

Answer 12

ohhhh okay okay

Answer 13

After that you can put your touch by replacing y and plug in its substitution that you have set at the beginning!

Answer 14

( \[(x^4+14) (x^4-13)\]

Answer 15

is it- ?

Answer 16

What I meant is to replace y with (x+4)
so if you replace y^2, you will get (x+4)^2
in other words, the final result will be
\[[(x+4)^2-2][(x+4)^2-3]\]
got it?

Answer 17

ahh yeah

Answer 18

Is it clear now?

Answer 19

im still not sure how to go on from there-

Answer 20

Where are you stuck?

Answer 21

i still dont really know how to find x

Answer 22

Oh sorry I thought that the question just to factor it.
Ok.
As you got the final result we have, can you equate every parentheses to zero?

Answer 23

Equate parentheses ?

Answer 24

\[(x+4)^2-3=0\]
and
\[(x+4)^2-2=0\]

Answer 25

ohhhh

Answer 26

I think you got it!

Answer 27

yeah yeah i do! Is it possible for you to help me with another question-

Answer 28

Of course, brother!
Why not?
Are you satisfied of that one firstly?

Answer 29

Yeah i managed to get the answers thnks !
Can you also help me w the question
Solve for x given \[(x^2-2x)^2 - 11(x^2-2x)+24=0\]

Answer 30

Can we start in new post, please?
I'd appreciate!

Answer 31

oky doky

Answer 32

Thanks