Menu Close

Find-all-solutions-x-to-the-equation-x-3-bx-2-cx-d-0-where-b-c-d-are-constants-from-C-

Tinku Tara 0 Comments
Pin




Question Number 3548 by Yozzii last updated on 15/Dec/15
Find all solutions x to the equation x^3 +bx^2 +cx+d=0 where b,c,d are constants from C.
$${Find}\:{all}\:{solutions}\:{x}\:{to}\:{the}\:{equation} \\ $$$${x}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0}\:{where}\:{b},{c},{d}\:{are}\: \\ $$$${constants}\:{from}\:\mathbb{C}.\: \\ $$$$ \\ $$
Answered by RasheedSindhi last updated on 15/Dec/15
Let b=b_1 +b_2 i c=c_1 +c_2 i d=d_1 +d_2 i where b_1 ,b_2 ,c_1 ,c_2 ,d_1 ,d_2 are all reals x^3 +b_1 x^2 +c_1 x+d_1 =0 ∧b_2 x^2 +c_2 x+d_2 =0
$${Let}\:{b}={b}_{\mathrm{1}} +{b}_{\mathrm{2}} {i} \\ $$$$\:\:\:\:\:\:\:\:{c}={c}_{\mathrm{1}} +{c}_{\mathrm{2}} {i} \\ $$$$\:\:\:\:\:\:\:\:{d}={d}_{\mathrm{1}} +{d}_{\mathrm{2}} {i} \\ $$$${where}\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,{c}_{\mathrm{1}} ,{c}_{\mathrm{2}} ,{d}_{\mathrm{1}} ,{d}_{\mathrm{2}} \:{are}\:{all}\:{reals} \\ $$$${x}^{\mathrm{3}} +{b}_{\mathrm{1}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{d}_{\mathrm{1}} =\mathrm{0}\:\wedge{b}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{2}} {x}+{d}_{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$
Commented by prakash jain last updated on 15/Dec/15
By substituting: (x^3 +b_1 x^2 +c_1 x+d_1 )+i(b_2 x^2 +c_2 x+d_2 )=0 It cannot be inferred that (b_2 x^2 +c_2 x+d_2 ) =0 unless it is given that x∈R. If x∈C, then (b_2 x^2 +c_2 x+d_2 )=u+v(√(−1))
$$\mathrm{By}\:\mathrm{substituting}: \\ $$$$\left({x}^{\mathrm{3}} +{b}_{\mathrm{1}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{d}_{\mathrm{1}} \right)+{i}\left({b}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{2}} {x}+{d}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{It}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{inferred}\:\mathrm{that}\: \\ $$$$\left({b}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{2}} {x}+{d}_{\mathrm{2}} \right)\:=\mathrm{0} \\ $$$$\mathrm{unless}\:\mathrm{it}\:\mathrm{is}\:\mathrm{given}\:\mathrm{that}\:{x}\in\mathbb{R}. \\ $$$$\mathrm{If}\:{x}\in\mathbb{C},\:\mathrm{then}\:\left({b}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{2}} {x}+{d}_{\mathrm{2}} \right)={u}+{v}\sqrt{−\mathrm{1}} \\ $$
Commented by RasheedSindhi last updated on 15/Dec/15
THαnkS for guidance!
$$\mathcal{TH}\alpha{n}\Bbbk\mathcal{S}\:{for}\:{guidance}! \\ $$$$ \\ $$
Answered by prakash jain last updated on 15/Dec/15
x=y−(b/3) (y−(b/3))^3 +b(y−(b/3))^2 +c(y−(b/3))+d=0 y^3 −by^2 +y(b^2 /3)−(b^3 /(27))+by^2 −2b^3 (y/3)+(b^3 /9)+cy−((bc)/3)+d=0 y^3 +y(c−(b^3 /3))+(d−((bc)/3)+((2b^3 )/(27)))=0 p=c−(b^3 /3), q=d−((bc)/3)+((2b^3 )/(27)) y^3 +py+q=0 subtitute y=u−v (u−v)^3 +p(u−v)+q=0 (q−(v^3 −u^3 ))+(u−v)(p−3uv)=0 u−v is a solution if q=v^3 −u^3 , p=3uv p=3uv⇒v=(p/(3u)) q=v^3 −u^3 ⇒q=(p^3 /(27u^3 ))−u^3 ....(A) (A) is quadratic in u^3 ⇒u can be found then we find v. so we know u−v is solution. once we know one solution of cubic then other 2 can be found after factorizing. There is also a cubic formula which is derived based on the above steps. But it is very hard even to type. The steps are easy to remember.
$${x}={y}−\frac{{b}}{\mathrm{3}} \\ $$$$\left({y}−\frac{{b}}{\mathrm{3}}\right)^{\mathrm{3}} +{b}\left({y}−\frac{{b}}{\mathrm{3}}\right)^{\mathrm{2}} +{c}\left({y}−\frac{{b}}{\mathrm{3}}\right)+{d}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} −{by}^{\mathrm{2}} +{y}\frac{{b}^{\mathrm{2}} }{\mathrm{3}}−\frac{{b}^{\mathrm{3}} }{\mathrm{27}}+{by}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{3}} \frac{{y}}{\mathrm{3}}+\frac{{b}^{\mathrm{3}} }{\mathrm{9}}+{cy}−\frac{{bc}}{\mathrm{3}}+{d}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} +{y}\left({c}−\frac{{b}^{\mathrm{3}} }{\mathrm{3}}\right)+\left({d}−\frac{{bc}}{\mathrm{3}}+\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}}\right)=\mathrm{0} \\ $$$${p}={c}−\frac{{b}^{\mathrm{3}} }{\mathrm{3}},\:{q}={d}−\frac{{bc}}{\mathrm{3}}+\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}} \\ $$$${y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$\mathrm{subtitute}\:{y}={u}−{v} \\ $$$$\left({u}−{v}\right)^{\mathrm{3}} +{p}\left({u}−{v}\right)+{q}=\mathrm{0} \\ $$$$\left({q}−\left({v}^{\mathrm{3}} −{u}^{\mathrm{3}} \right)\right)+\left({u}−{v}\right)\left({p}−\mathrm{3}{uv}\right)=\mathrm{0} \\ $$$${u}−{v}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{if}\:{q}={v}^{\mathrm{3}} −{u}^{\mathrm{3}} ,\:{p}=\mathrm{3}{uv} \\ $$$${p}=\mathrm{3}{uv}\Rightarrow{v}=\frac{{p}}{\mathrm{3}{u}} \\ $$$${q}={v}^{\mathrm{3}} −{u}^{\mathrm{3}} \Rightarrow{q}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}{u}^{\mathrm{3}} }−{u}^{\mathrm{3}} \:\:\:\:….\left({A}\right) \\ $$$$\left({A}\right)\:\mathrm{is}\:\mathrm{quadratic}\:\mathrm{in}\:{u}^{\mathrm{3}} \Rightarrow{u}\:{can}\:{be}\:{found} \\ $$$${then}\:{we}\:{find}\:{v}. \\ $$$${so}\:{we}\:{know}\:{u}−{v}\:{is}\:{solution}. \\ $$$${once}\:{we}\:{know}\:{one}\:{solution}\:{of}\:{cubic}\:{then} \\ $$$${other}\:\mathrm{2}\:{can}\:{be}\:{found}\:{after}\:{factorizing}. \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{cubic}\:\mathrm{formula}\:\mathrm{which}\:\mathrm{is}\:\mathrm{derived} \\ $$$$\mathrm{based}\:\mathrm{on}\:\mathrm{the}\:\mathrm{above}\:\mathrm{steps}.\:\mathrm{But}\:\mathrm{it}\:\mathrm{is}\:\mathrm{very}\:\mathrm{hard} \\ $$$$\mathrm{even}\:\mathrm{to}\:\mathrm{type}.\:\mathrm{The}\:\mathrm{steps}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{remember}. \\ $$
Commented by Yozzii last updated on 15/Dec/15
Thanks a lot. I′m wondering how one creates such methods because I′m interested now in solving a good few of these polynomials, perhaps up to degree 10 (being terribly ambitious).
$${Thanks}\:{a}\:{lot}.\:{I}'{m}\:{wondering}\:{how} \\ $$$${one}\:{creates}\:{such}\:{methods}\:{because} \\ $$$${I}'{m}\:{interested}\:{now}\:{in}\:{solving}\: \\ $$$${a}\:{good}\:{few}\:{of}\:{these}\:{polynomials}, \\ $$$${perhaps}\:{up}\:{to}\:{degree}\:\mathrm{10}\:\left({being}\:\right. \\ $$$$\left.{terribly}\:{ambitious}\right). \\ $$$$ \\ $$
Commented by prakash jain last updated on 15/Dec/15
Abel Ruffini Theorem: There is no general algebraic solution for polynomial of degree 5 or higher with arbitrary coefficients. So you cannot find a formula. Formula upto degree 4 are already available.
$$\mathrm{Abel}\:\mathrm{Ruffini}\:\mathrm{Theorem}:\:\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{general} \\ $$$$\mathrm{algebraic}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{polynomial}\:\mathrm{of}\:\mathrm{degree} \\ $$$$\mathrm{5}\:\mathrm{or}\:\mathrm{higher}\:\mathrm{with}\:\mathrm{arbitrary}\:\mathrm{coefficients}. \\ $$$$\mathrm{So}\:\mathrm{you}\:\mathrm{cannot}\:\mathrm{find}\:\mathrm{a}\:\mathrm{formula}. \\ $$$$\mathrm{Formula}\:\mathrm{upto}\:\mathrm{degree}\:\mathrm{4}\:\mathrm{are}\:\mathrm{already}\:\mathrm{available}. \\ $$
Commented by Yozzii last updated on 15/Dec/15
I see. It′d be good to learn the proofs for degree 4.
$${I}\:{see}.\:{It}'{d}\:{be}\:{good}\:{to}\:{learn}\:{the}\:{proofs} \\ $$$${for}\:{degree}\:\mathrm{4}. \\ $$
Commented by Yozzii last updated on 15/Dec/15
After degree 4 we′d rely on numerical methods only to solve polynomials?
$${After}\:{degree}\:\mathrm{4}\:{we}'{d}\:{rely}\:{on}\: \\ $$$${numerical}\:{methods}\:{only}\:{to}\:{solve} \\ $$$${polynomials}? \\ $$
Commented by prakash jain last updated on 15/Dec/15
For a general equation with arbitrary coefficients only numeric solution. However there are equation of degree 5 which can be solved if coefficients meet certain criteria. So for your further study you may want to work on solvable equation and extending the result wherever you can.
$$\mathrm{For}\:\mathrm{a}\:\mathrm{general}\:\mathrm{equation}\:\mathrm{with}\:\mathrm{arbitrary}\:\mathrm{coefficients} \\ $$$$\mathrm{only}\:\mathrm{numeric}\:\mathrm{solution}.\:\mathrm{However}\:\mathrm{there}\:\mathrm{are} \\ $$$$\mathrm{equation}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{5}\:\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$$$\mathrm{if}\:\mathrm{coefficients}\:\mathrm{meet}\:\mathrm{certain}\:\mathrm{criteria}. \\ $$$$\mathrm{So}\:\mathrm{for}\:\mathrm{your}\:\mathrm{further}\:\mathrm{study}\:\mathrm{you}\:\mathrm{may}\:\mathrm{want}\:\mathrm{to} \\ $$$$\mathrm{work}\:\mathrm{on}\:\mathrm{solvable}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{extending} \\ $$$$\mathrm{the}\:\mathrm{result}\:\mathrm{wherever}\:\mathrm{you}\:\mathrm{can}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *