anonymous
  • anonymous
please help!; how do you find the solutions for x^4-x^2-12=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Factor it out!
anonymous
  • anonymous
(x^2 - 4)(x^2 + 3) = 0
anonymous
  • anonymous
i did and got 0,3,-4

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

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