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dddan

  • 2 years ago

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

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  1. tejasree
    • 2 years ago
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    Factor it out!

  2. tejasree
    • 2 years ago
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    (x^2 - 4)(x^2 + 3) = 0

  3. dddan
    • 2 years ago
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    i did and got 0,3,-4

  4. dddan
    • 2 years ago
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    i just want to see if im right

  5. tejasree
    • 2 years ago
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    no!

  6. tejasree
    • 2 years ago
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    x^2 - 4 = 0 (x-2)(x+2) = 0 x=2,-2

  7. cwrw238
    • 2 years ago
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    let y = x^2 then y^2 - y -12 = 0 can you solve this?

  8. tejasree
    • 2 years ago
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    wait there are more solutions

  9. cwrw238
    • 2 years ago
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    yup there will be 4 solution because the degree of the equation is 4

  10. tejasree
    • 2 years ago
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    x^2 + 3 = 0 x^2 = -3 \[x = i \sqrt{3}, -i \sqrt{3}\]

  11. tejasree
    • 2 years ago
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    the solutions are: \[x = 2, -2, i \sqrt{3}, -i \sqrt{3}\]

  12. tejasree
    • 2 years ago
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    am I correct @cwrw238

  13. dddan
    • 2 years ago
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    i understand why there 4 solutions but where did you get them from?

  14. tejasree
    • 2 years ago
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    First we factored them out, didnt we? then we set each factor equal to zero

  15. jim_thompson5910
    • 2 years ago
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    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

  16. jim_thompson5910
    • 2 years ago
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    solve z^2-z-12=0 and keep in mind that z = x^2 so you can solve for x

  17. dddan
    • 2 years ago
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    o because it x^2=3 its the square root plus i right

  18. dddan
    • 2 years ago
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    ok i got it - thank you so much guys

  19. tejasree
    • 2 years ago
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    there you go @jim_thompson5910 :)

  20. tejasree
    • 2 years ago
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    thank you! :)

  21. tejasree
    • 2 years ago
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    this site is really helpful: mathway.com

  22. cwrw238
    • 2 years ago
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    @tegasree - u r correct

  23. tejasree
    • 2 years ago
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    thank you! :)

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