Question

please help!;
how do you find the solutions for x^4-x^2-12=0

Answer 1

Factor it out!

Answer 2

(x^2 - 4)(x^2 + 3) = 0

Answer 3

i did and got 0,3,-4

Answer 4

i just want to see if im right

Answer 5

no!

Answer 6

x^2 - 4 = 0
(x-2)(x+2) = 0
x=2,-2

Answer 7

let y = x^2
then y^2 - y -12 = 0
can you solve this?

Answer 8

wait there are more solutions

Answer 9

yup there will be 4 solution because the degree of the equation is 4

Answer 10

x^2 + 3 = 0
x^2 = -3
\[x = i \sqrt{3}, -i \sqrt{3}\]

Answer 11

the solutions are:
\[x = 2, -2, i \sqrt{3}, -i \sqrt{3}\]

Answer 12

am I correct @cwrw238

Answer 13

i understand why there 4 solutions but where did you get them from?

Answer 14

First we factored them out, didnt we?
then we set each factor equal to zero

Answer 15

It might be easier to let z = x^2

so z^2 = x^4

which allows us to go from

x^4-x^2-12=0

to

z^2-z-12=0

Answer 16

solve z^2-z-12=0 and keep in mind that z = x^2 so you can solve for x

Answer 17

o because it x^2=3 its the square root plus i right

Answer 18

ok i got it - thank you so much guys

Answer 19

there you go @jim_thompson5910 :)

Answer 20

thank you! :)

Answer 21

this site is really helpful:
mathway.com

Answer 22

@tegasree - u r correct

Answer 23

thank you! :)