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jaydemarie4

  • 3 years ago

What are the real or imaginary solutions of the polynomial equation?

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  1. rahul91
    • 3 years ago
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    put x^4=t^2 and try to solve

  2. jaydemarie4
    • 3 years ago
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    i still don't understand..

  3. rahul91
    • 3 years ago
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    substitute x^4 = t^2 you get t^2 - 41t + 400 = 0 can u solve this quadratic equation

  4. jaydemarie4
    • 3 years ago
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    i can try i just dont understand algebra at all

  5. rahul91
    • 3 years ago
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    just solve for t then put x^2 = t. u get x. u got it ? don't worry keep on practising and you will start loving the subject

  6. zordoloom
    • 3 years ago
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    BTW, there are 4 solutions and all of them are real. No imaginary zeros

  7. jaydemarie4
    • 3 years ago
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    thanks you guys i will try to figure it out!

  8. anonymous
    • 3 years ago
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    try factoring \[t^2-41t+400=0\] it actually factors

  9. anonymous
    • 3 years ago
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    and a hint is that the two numbers whose product is 400 are both perfect squares

  10. jaydemarie4
    • 3 years ago
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    thank you very much for your help!

  11. anonymous
    • 3 years ago
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    did you get \((t-16)(t-25)=0\)? if so then you go to \(x^2-16=0,x^2-25=0\) and solve both of those for \(x\)

  12. jaydemarie4
    • 3 years ago
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    oh alrighty. thanks :)

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