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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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