Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

UnkleRhaukus

  • 3 years ago

roots of a trinomial \[ax^3+bx^2+cx+d=0\] let the roots be \(p,q,r\) When \(x\) is equal a root the equation equals zero \[(x-p)(x-q)(x-r)=0\] \[(x-p)(x^2-(q+r)x+qr)=0\] \[x\left(x^2-(q+r)x+qr\right)-p\left(x^2-(q+r)x+qr\right)=0\] \[\left(x^3-(q+r)x^2+(qr)x\right)-\left(px^2-p(q+r)x-qpr\right)=0\] \[x^3-(p+q+r)x^2+(qr+pq+pr)x-pqr=0\] comparing with \[x^3+\frac bax^2+\frac cax+\frac da=0\] \[-(p+q+r)=\frac ba\qquad\text{the sum of the roots is } -\frac ba\] \[(qr+pq+pr)=\frac ca\qquad\text{sum of products of pairs of the roots is } \frac ca\] \[-pqr=\frac da\qquad \text{the product of the roots is} -\frac da \]

  • This Question is Closed
  1. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351834318182:dw|

  2. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351834441300:dw|

  3. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I divided to a.

  4. UnkleRhaukus
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    what about complex solutions

  5. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351834556416:dw|

  6. UnkleRhaukus
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i guess a trinomial always has at least one real root

  7. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351834728740:dw|

  8. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351834848416:dw|

  9. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if the trinomial has real co-efficients, then complex roots occur in pairs, hence it must have atleast 1 real solution .

  10. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351834939890:dw|

  11. UnkleRhaukus
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i see

  12. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351835035013:dw|

  13. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351835192802:dw|

  14. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351835381933:dw|

  15. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    this is the general for all n-order polynomial equation.

  16. UnkleRhaukus
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    awesome

  17. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    notice the sign change everytime, product of all roots = \( (-1)^n a_0/a_n \)

  18. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351835549510:dw|

  19. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I should mention that there are all real and complex roots in formulas.

  20. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351839125496:dw|

  21. mahmit2012
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1351839193349:dw|

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy