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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
\[(xp)(xq)(xr)=0\]
\[(xp)(x^2(q+r)x+qr)=0\]
\[x\left(x^2(q+r)x+qr\right)p\left(x^2(q+r)x+qr\right)=0\]
\[\left(x^3(q+r)x^2+(qr)x\right)\left(px^2p(q+r)xqpr\right)=0\]
\[x^3(p+q+r)x^2+(qr+pq+pr)xpqr=0\]
comparing with
\[x^3+\frac bax^2+\frac cax+\frac da=0\]
\[(p+q+r)=\frac ba\qquad\text{the sum of the roots is } \frac ba\]
\[(qr+pq+pr)=\frac ca\qquad\text{sum of products of pairs of the roots is } \frac ca\]
\[pqr=\frac da\qquad \text{the product of the roots is} \frac da \]
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 \[(xp)(xq)(xr)=0\] \[(xp)(x^2(q+r)x+qr)=0\] \[x\left(x^2(q+r)x+qr\right)p\left(x^2(q+r)x+qr\right)=0\] \[\left(x^3(q+r)x^2+(qr)x\right)\left(px^2p(q+r)xqpr\right)=0\] \[x^3(p+q+r)x^2+(qr+pq+pr)xpqr=0\] comparing with \[x^3+\frac bax^2+\frac cax+\frac da=0\] \[(p+q+r)=\frac ba\qquad\text{the sum of the roots is } \frac ba\] \[(qr+pq+pr)=\frac ca\qquad\text{sum of products of pairs of the roots is } \frac ca\] \[pqr=\frac da\qquad \text{the product of the roots is} \frac da \]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351834318182:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351834441300:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2what about complex solutions

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351834556416:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.2i guess a trinomial always has at least one real root

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351834728740:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351834848416:dw

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0if the trinomial has real coefficients, then complex roots occur in pairs, hence it must have atleast 1 real solution .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351834939890:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351835035013:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351835192802:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351835381933:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is the general for all norder polynomial equation.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0notice the sign change everytime, product of all roots = \( (1)^n a_0/a_n \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351835549510:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I should mention that there are all real and complex roots in formulas.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351839125496:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351839193349:dw
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