Question

x^4 - 1= 0

Answer 1

\[x^4-1=0\]\[x^4=1\]\[x=+/-1\]

Answer 2

close stamp, but you're missing 2 roots

Answer 3

oh really its not the square root of anything?

Answer 4

\[(x^2+1)(x^2-1)\]\[(x^2+1)(x-1)(x+1)\]

Answer 5

\[a^2-b^2=(a+b)(a-b)\]

in this case

\[(x^2)^2-1= (x^2-1)(x^2+1)\]

Answer 6

then x^2-1 can be factored out further

then solve for x

Answer 7

yeah that's what I thought x = four muliplide by the square root root of = or - 1

Answer 8

= or - 1

Answer 9

there are 2 other answers

x^2+1=0

x^2=-1
x= +/- i

Answer 10

althought its imaginary, its still considered a root

Answer 11

yeah sorry typo

Answer 12

so it is an imaginary number?

Answer 13

ok starting from the beginning

we have \[x^4-1 =0\]
we want to solve for x
so we first need to factor the expression on the left completely

\[x^4-1 =(x^2-1)(x^2+1)=0\]

now using the zeroproperty of multiplication means that either \[x^2-1 =0\] and/or \[x^2+1 =0\]

so now we need to solve for 2 equations,

starting with \[x^2-1=0\]
\[x^2-1= (x+1)(x-1)=0\]

now again using zero property of multiplication, we get x-1=0 and x+1=0
solving for x would give us
x=1 and x=-1

those are 2 roots, 2of the answers

now we still have to solve for \[x^2+1=0\]

\[x^2=-1\]
\[x=\pm \sqrt{-1}\]

now \[\sqrt{-1}=i\]
where i is an imaginary number

so that means the other 2 answers/roots are x=i and x=-i

Answer 14

long story short, there are 4 different answers

Answer 15

cool thank you very much