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

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

Mathematics
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Factor it out!
(x^2 - 4)(x^2 + 3) = 0
i did and got 0,3,-4

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i just want to see if im right
no!
x^2 - 4 = 0 (x-2)(x+2) = 0 x=2,-2
let y = x^2 then y^2 - y -12 = 0 can you solve this?
wait there are more solutions
yup there will be 4 solution because the degree of the equation is 4
x^2 + 3 = 0 x^2 = -3 \[x = i \sqrt{3}, -i \sqrt{3}\]
the solutions are: \[x = 2, -2, i \sqrt{3}, -i \sqrt{3}\]
am I correct @cwrw238
i understand why there 4 solutions but where did you get them from?
First we factored them out, didnt we? then we set each factor equal to zero
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
solve z^2-z-12=0 and keep in mind that z = x^2 so you can solve for x
o because it x^2=3 its the square root plus i right
ok i got it - thank you so much guys
there you go @jim_thompson5910 :)
thank you! :)
this site is really helpful: mathway.com
@tegasree - u r correct
thank you! :)

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