UnkleRhaukus
  • UnkleRhaukus
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 \[x^3+\frac bax^2+\frac cax+\frac da=0\] \[-(p+q+r)=\frac ba\qquad\text{the sum of the roots is } -\frac ba\] \[(qr+pq+pr)=\frac ca\qquad\text{sum of products of pairs of the roots is } \frac ca\] \[-pqr=\frac da\qquad \text{the product of the roots is} -\frac da \]
Algebra
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1351834318182:dw|
anonymous
  • anonymous
|dw:1351834441300:dw|
anonymous
  • anonymous
I divided to a.

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More answers

UnkleRhaukus
  • UnkleRhaukus
what about complex solutions
anonymous
  • anonymous
|dw:1351834556416:dw|
UnkleRhaukus
  • UnkleRhaukus
i guess a trinomial always has at least one real root
anonymous
  • anonymous
|dw:1351834728740:dw|
anonymous
  • anonymous
|dw:1351834848416:dw|
hartnn
  • hartnn
if the trinomial has real co-efficients, then complex roots occur in pairs, hence it must have atleast 1 real solution .
anonymous
  • anonymous
|dw:1351834939890:dw|
UnkleRhaukus
  • UnkleRhaukus
i see
anonymous
  • anonymous
|dw:1351835035013:dw|
anonymous
  • anonymous
|dw:1351835192802:dw|
anonymous
  • anonymous
|dw:1351835381933:dw|
anonymous
  • anonymous
this is the general for all n-order polynomial equation.
UnkleRhaukus
  • UnkleRhaukus
awesome
hartnn
  • hartnn
notice the sign change everytime, product of all roots = \( (-1)^n a_0/a_n \)
anonymous
  • anonymous
|dw:1351835549510:dw|
anonymous
  • anonymous
I should mention that there are all real and complex roots in formulas.
anonymous
  • anonymous
|dw:1351839125496:dw|
anonymous
  • anonymous
|dw:1351839193349:dw|

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