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mskyeg

  • 2 years ago

Find the zeros of the function. State the multiplicity of any multiple zeros. y=x^4-8x^2+16

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  1. hartnn
    • 2 years ago
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    have u tried this ?

  2. mskyeg
    • 2 years ago
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    ya im supposed to find the GCF but i don't think there is one so i dont know what to do

  3. hartnn
    • 2 years ago
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    GCF ? first put x^2=y what u get ?

  4. mskyeg
    • 2 years ago
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    why would x^2 equal y

  5. hartnn
    • 2 years ago
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    oh, it was just for simplification, u can put x^2 =t .... or z ..

  6. hartnn
    • 2 years ago
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    i didn't mean that y on left side...

  7. mskyeg
    • 2 years ago
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    square root both sides

  8. hartnn
    • 2 years ago
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    to find zeros of y=x^4-8x^2+16 u put y=0 x^4-8x^2+16=0 now if we put x^2 = t t^2 -8t +16 = 0 this is quadratic in t, can u solve it ?

  9. mskyeg
    • 2 years ago
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    so just use the quadratic formula

  10. hartnn
    • 2 years ago
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    or u can factor it .

  11. mskyeg
    • 2 years ago
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    the answer is -4 or 4 but the answer is supposed to be -2 or 2

  12. mskyeg
    • 2 years ago
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    how do i factor it

  13. hartnn
    • 2 years ago
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    yes, because x^2 is 4,4

  14. hartnn
    • 2 years ago
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    t^2-8t+16 = 0 t^2 -4t -4t +16 = 0

  15. hartnn
    • 2 years ago
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    ok?

  16. hartnn
    • 2 years ago
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    what u got t as ?

  17. mskyeg
    • 2 years ago
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    im sorry what?

  18. hartnn
    • 2 years ago
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    t^2 -4t -4t +16 = 0 can u solve for t ?

  19. mskyeg
    • 2 years ago
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    no thats why i asked

  20. hartnn
    • 2 years ago
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    t^2 -4t -4t +16 = 0 factor out t from first 2 terms , what u get ?

  21. mskyeg
    • 2 years ago
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    i mean t(t-8)+16=0

  22. hartnn
    • 2 years ago
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    t^2 -4t -4t +16 = 0 from this.... factor out t from 1st 2 terms....

  23. mskyeg
    • 2 years ago
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    thats what i thought i just did

  24. hartnn
    • 2 years ago
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    u factored t from t^2 -8t +16 = 0 i asked u to factor out t from t^2 -4t -4t +16 = 0

  25. mskyeg
    • 2 years ago
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    t(t-4-4)+16=0

  26. hartnn
    • 2 years ago
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    i mentioned 'from first two terms' ...

  27. mskyeg
    • 2 years ago
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    t(t-4-4t)+16=0

  28. hartnn
    • 2 years ago
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    t^2-4t-4t+16=0 t(t-4) -4 (t-4) = 0 did u get this ?

  29. mskyeg
    • 2 years ago
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    oh ok that makes sense

  30. hartnn
    • 2 years ago
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    t(t-4) -4 (t-4) = 0 (t-4)(t-4) =0 i factored t-4 from that, got this ?

  31. mskyeg
    • 2 years ago
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    how did u get rid of the t and -4 ouside the parentheses

  32. hartnn
    • 2 years ago
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    i did not get rid of that...its in the 2nd brackets '()' ... (t-4)

  33. hartnn
    • 2 years ago
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    if u are getting confused let a=t-4 t(t-4) -4 (t-4) = 0 ta-4a = 0 a(t-4)=0 now put back a= t-4 (t-4)(t-4) = 0 got this ?

  34. mskyeg
    • 2 years ago
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    ok

  35. hartnn
    • 2 years ago
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    so, t=4,4 which gives u x^2 = 4, 4

  36. hartnn
    • 2 years ago
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    so whats x = ?

  37. mskyeg
    • 2 years ago
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    x=2 and -2

  38. mskyeg
    • 2 years ago
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    x=+or - the square root of 4

  39. hartnn
    • 2 years ago
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    yes thats correct...x=2,-2 for x^2 =4 now since x^2 = 4,4 ....u get x= 2,2,-2,-2 that is a root =2 with muliplicity 2 and another root =-2 with multiplicity = 2

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