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ax-3-bx-2-cx-d-0-pls-solve-




Question Number 11770 by Nayon last updated on 31/Mar/17
ax^3 +bx^2 +cx+d=0 pls.solve
$${ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0}\:{pls}.{solve} \\ $$
Answered by Joel576 last updated on 31/Mar/17
f(x) = x^3  + (b/a)x^2  + (c/a)x + (d/a) = 0   ... (i)  Let p, q, r are the roots from f(x)  We get:  f(x) = (x − p)(x − q)(x − r) = 0            = (x^2  − qx − px + pq)(x − r) = 0            = x^3  − rx^2  − qx^2  + qrx − px^2  + rpx + pqx − pqr            = x^3  − (p + q + r)x^2  + (pq + qr + rp)x − pqr = 0   ... (ii)  From (i) and (ii) we can conclude:  p + q + r = −(b/a)  pq + qr + rp = (c/a)  pqr = −(d/a)
$${f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:\frac{{b}}{{a}}{x}^{\mathrm{2}} \:+\:\frac{{c}}{{a}}{x}\:+\:\frac{{d}}{{a}}\:=\:\mathrm{0}\:\:\:…\:\left({i}\right) \\ $$$$\mathrm{Let}\:{p},\:{q},\:{r}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{from}\:{f}\left({x}\right) \\ $$$$\mathrm{We}\:\mathrm{get}: \\ $$$${f}\left({x}\right)\:=\:\left({x}\:−\:{p}\right)\left({x}\:−\:{q}\right)\left({x}\:−\:{r}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\left({x}^{\mathrm{2}} \:−\:{qx}\:−\:{px}\:+\:{pq}\right)\left({x}\:−\:{r}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{x}^{\mathrm{3}} \:−\:{rx}^{\mathrm{2}} \:−\:{qx}^{\mathrm{2}} \:+\:{qrx}\:−\:{px}^{\mathrm{2}} \:+\:{rpx}\:+\:{pqx}\:−\:{pqr} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{x}^{\mathrm{3}} \:−\:\left({p}\:+\:{q}\:+\:{r}\right){x}^{\mathrm{2}} \:+\:\left({pq}\:+\:{qr}\:+\:{rp}\right){x}\:−\:{pqr}\:=\:\mathrm{0}\:\:\:…\:\left({ii}\right) \\ $$$$\mathrm{From}\:\left({i}\right)\:\mathrm{and}\:\left({ii}\right)\:\mathrm{we}\:\mathrm{can}\:\mathrm{conclude}: \\ $$$${p}\:+\:{q}\:+\:{r}\:=\:−\frac{{b}}{{a}} \\ $$$${pq}\:+\:{qr}\:+\:{rp}\:=\:\frac{{c}}{{a}} \\ $$$${pqr}\:=\:−\frac{{d}}{{a}} \\ $$
Commented by Nayon last updated on 31/Mar/17
can i get the roots by solving thosr  equations??
$${can}\:{i}\:{get}\:{the}\:{roots}\:{by}\:{solving}\:{thosr} \\ $$$${equations}?? \\ $$
Commented by Joel576 last updated on 31/Mar/17
Commented by Nayon last updated on 31/Mar/17
how can i get p,q,r?
$${how}\:{can}\:{i}\:{get}\:{p},{q},{r}? \\ $$$$ \\ $$
Commented by mrW1 last updated on 31/Mar/17
To solve the equation system for p,  q, r you also need to solve an other cubic  equation, so you reach nothing in  this way.     To be honest, we don′t need  to reinvent the wheel again. The   general way to solve a cubic equation of  the form ax^3 +bx^2 +cx+d=0 is known  since hundreds of years. People have  created general formulas like those above  which we can use directly. The way  of solution is too long to be given here.  In Google you can find a lot of stuff  about it. So far as I know, the basic  idea is to do some substitutions to put  the equation into a new form whose  solution is known.  I can give you an interesting link  which contains a detailed description.
$${To}\:{solve}\:{the}\:{equation}\:{system}\:{for}\:{p}, \\ $$$${q},\:{r}\:{you}\:{also}\:{need}\:{to}\:{solve}\:{an}\:{other}\:{cubic} \\ $$$${equation},\:{so}\:{you}\:{reach}\:{nothing}\:{in} \\ $$$${this}\:{way}.\: \\ $$$$ \\ $$$${To}\:{be}\:{honest},\:{we}\:{don}'{t}\:{need} \\ $$$${to}\:{reinvent}\:{the}\:{wheel}\:{again}.\:{The}\: \\ $$$${general}\:{way}\:{to}\:{solve}\:{a}\:{cubic}\:{equation}\:{of} \\ $$$${the}\:{form}\:{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0}\:{is}\:{known} \\ $$$${since}\:{hundreds}\:{of}\:{years}.\:{People}\:{have} \\ $$$${created}\:{general}\:{formulas}\:{like}\:{those}\:{above} \\ $$$${which}\:{we}\:{can}\:{use}\:{directly}.\:{The}\:{way} \\ $$$${of}\:{solution}\:{is}\:{too}\:{long}\:{to}\:{be}\:{given}\:{here}. \\ $$$${In}\:{Google}\:{you}\:{can}\:{find}\:{a}\:{lot}\:{of}\:{stuff} \\ $$$${about}\:{it}.\:{So}\:{far}\:{as}\:{I}\:{know},\:{the}\:{basic} \\ $$$${idea}\:{is}\:{to}\:{do}\:{some}\:{substitutions}\:{to}\:{put} \\ $$$${the}\:{equation}\:{into}\:{a}\:{new}\:{form}\:{whose} \\ $$$${solution}\:{is}\:{known}. \\ $$$${I}\:{can}\:{give}\:{you}\:{an}\:{interesting}\:{link} \\ $$$${which}\:{contains}\:{a}\:{detailed}\:{description}. \\ $$
Commented by mrW1 last updated on 31/Mar/17

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