Question

X^4=2x^2-1
solve algebracally

Answer 1

\[x^4-2x^2+1=0\]
\[(x^2)^2-2x^2+1=0\]
replace x^2 with u
and see if you can figure out now
let me know if you still have trouble after replacing with u

Answer 2

well then if you use the quadratic form its not real is that correcr?

Answer 3

\[u^2-2u+1=0\]
this is factorable right?
\[(u-1)(u-1)=0\]

Answer 4

u-1=0 =>u=1
but u=x^2
x^2=1=>x=1 or x=-1

Answer 5

so one more thing to ask for factoring what number am I factoring

Answer 6

you are factoring the polynomial u^2-2u+1
you can use quadratic and will give you one real solution

Answer 7

quadratic formula*

Answer 8

yeah yeah, but where did you get the 1 and 1 from (u−1)(u−1)=0

Answer 9

\[u=\frac{2 \pm \sqrt{(-2)^2-4(1)(1)}}{2(1)}=\frac{2 \pm \sqrt{4-4}}{2}=\frac{2 \pm 0}{2}=\frac{2}{2}=1\]

Answer 10

ok got it, thanks

Answer 11

I used the following process to factor:
u^2+bu+c
only works when coefficient of u^2 is 1
find factors of c that add up to be b
then put the numbers in the blanks below
(u+__)(u+__)

Answer 12

u=2±(−2)2−4(1)(1)−−−−−−−−−−−−√2(1)=2±4−4−−−−√2=2±02=22=1

Answer 13

at the second part how did you get I got -8 under the radical how did you get 0

Answer 14

(-2)^2=(-2)(-2)=4
4-4=0

Answer 15

\[u=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 16

oh ok never mind

Answer 17

i think you forgot to square the negative part
thats why you got -8