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UnkleRhaukus
Group Title
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
\[x^3+\frac bax^2+\frac cax+\frac da=0\]
\[(p+q+r)=\frac ba\qquad\text{the sum of the roots is } \frac ba\]
\[(qr+pq+pr)=\frac ca\qquad\text{sum of products of pairs of the roots is } \frac ca\]
\[pqr=\frac da\qquad \text{the product of the roots is} \frac da \]
 one year ago
 one year ago
UnkleRhaukus Group Title
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 \[x^3+\frac bax^2+\frac cax+\frac da=0\] \[(p+q+r)=\frac ba\qquad\text{the sum of the roots is } \frac ba\] \[(qr+pq+pr)=\frac ca\qquad\text{sum of products of pairs of the roots is } \frac ca\] \[pqr=\frac da\qquad \text{the product of the roots is} \frac da \]
 one year ago
 one year ago

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mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351834318182:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351834441300:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
I divided to a.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
what about complex solutions
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351834556416:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
i guess a trinomial always has at least one real root
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351834728740:dw
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mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351834848416:dw
 one year ago

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

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351834939890:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
i see
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351835035013:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351835192802:dw
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mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351835381933:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
this is the general for all norder polynomial equation.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.2
awesome
 one year ago

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

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351835549510:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
I should mention that there are all real and complex roots in formulas.
 one year ago

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dw:1351839125496:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.2
dw:1351839193349:dw
 one year ago
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