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Question Number 42826 by MJS last updated on 03/Sep/18

solving ax^4 +bx^3 +cx^2 +dx+e=0 (a≠0, b, c, d, e)∈Q special cases (easy to solve) ax^4 +e=0 solve at^2 +e=0 ⇒ x=±(√t_(1, 2) ) ax^4 +cx^2 +e=0 solve at^2 +ct+e=0 ⇒ x=±(√t_(1, 2) ) always try all factors of ±e because a(x−α)(x−β)(x−γ)(x−δ)=ax^4 +...+αβγδ ⇒ e=αβγδ next we must find the nature of the solutions 4 real solutions 2 real & 2 complex solutions 4 complex solutions a, b, c, d, e ∈Q ⇒ complex solutions always in conjugated pairs draw the function or calculate some values to find the number of real solutions divide by a x^4 +px^3 +qx^2 +rx+s=0 [p=(b/a) q=(c/a) r=(d/a) s=(e/a)] I′ll soon post some cases I′ve been able to solve as comments

$$\mathrm{solving} \\ $$$${ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e}=\mathrm{0} \\ $$$$\left({a}\neq\mathrm{0},\:{b},\:{c},\:{d},\:{e}\right)\in\mathbb{Q} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{special}\:\mathrm{cases}\:\left(\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{ax}^{\mathrm{4}} +{e}=\mathrm{0}\:\mathrm{solve}\:{at}^{\mathrm{2}} +{e}=\mathrm{0}\:\Rightarrow\:{x}=\pm\sqrt{{t}_{\mathrm{1},\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:{ax}^{\mathrm{4}} +{cx}^{\mathrm{2}} +{e}=\mathrm{0}\:\mathrm{solve}\:{at}^{\mathrm{2}} +{ct}+{e}=\mathrm{0}\:\Rightarrow\:{x}=\pm\sqrt{{t}_{\mathrm{1},\:\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{always}\:\mathrm{try}\:\mathrm{all}\:\mathrm{factors}\:\mathrm{of}\:\pm{e} \\ $$$$\mathrm{because}\:{a}\left({x}−\alpha\right)\left({x}−\beta\right)\left({x}−\gamma\right)\left({x}−\delta\right)={ax}^{\mathrm{4}} +...+\alpha\beta\gamma\delta \\ $$$$\Rightarrow\:{e}=\alpha\beta\gamma\delta \\ $$$$ \\ $$$$\mathrm{next}\:\mathrm{we}\:\mathrm{must}\:\mathrm{find}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solutions} \\ $$$$\mathrm{4}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\mathrm{2}\:\mathrm{real}\:\&\:\mathrm{2}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\mathrm{4}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$${a},\:{b},\:{c},\:{d},\:{e}\:\in\mathbb{Q}\:\Rightarrow\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{always}\:\mathrm{in} \\ $$$$\mathrm{conjugated}\:\mathrm{pairs} \\ $$$$\mathrm{draw}\:\mathrm{the}\:\mathrm{function}\:\mathrm{or}\:\mathrm{calculate}\:\mathrm{some}\:\mathrm{values} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$ \\ $$$$\mathrm{divide}\:\mathrm{by}\:{a} \\ $$$${x}^{\mathrm{4}} +{px}^{\mathrm{3}} +{qx}^{\mathrm{2}} +{rx}+{s}=\mathrm{0} \\ $$$$\left[{p}=\frac{{b}}{{a}}\:\:{q}=\frac{{c}}{{a}}\:\:{r}=\frac{{d}}{{a}}\:\:{s}=\frac{{e}}{{a}}\right] \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{soon}\:\mathrm{post}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{able}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{as}\:\mathrm{comments} \\ $$

Commented by Tawa1 last updated on 03/Sep/18

Great sir. God bless you sir

$$\mathrm{Great}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Necxx last updated on 03/Sep/18

Write a ten digit number containing 0-9. Situations are, * Each numbers should have used exactly once. * First n numbers should be divisible by n. e.g, 123, this is three digit number and divisible by 3. 1024, its a four digit number and divisible by 4....

Commented by MJS last updated on 03/Sep/18

4 real solutions x_1 , x_2 , x_3 , x_4 can be put as this: x_1 =α−(√β) x_2 =α+(√β) x_3 =γ−(√δ) x_4 =γ+(√δ) expanding (x−x_1 )(x−x_2 )(x−x_3 )(x−x_4 ) we get x^4 − −2(α+γ)x^3 + +(α^2 +4αγ−β+γ^2 −δ)x^2 − −2(α^2 γ+αγ^2 −αδ−βγ)x+ +(α^2 −β)(γ^2 −δ) this must be the same as x^4 +px^3 +qx^2 +rx+s ⇒ ⇒ { (((1) −2(α+γ)=p)),(((2) (α^2 +4αγ−β+γ^2 −δ)=q)),(((3) −2(α^2 γ+αγ^2 −αδ−βγ)=r)),(((4) (α^2 −β)(γ^2 −δ)=s)) :} solve (1) for α, (2) for β, (3) for δ, insert in (4) which leads to γ^6 +((3p)/2)γ^5 +((3p^2 +2q)/4)γ^4 +((p(p^2 +4q))/8)γ^3 +((2p^2 q+pr+q^2 −4s)/(16))γ^2 +((p(pr+q^2 −4s))/(32))γ−((p^2 s−pqr+r^2 )/(64))=0 γ=u−(p/4) u^6 −((3p^2 −8q)/(16))u^4 +((3p^4 −16(p^2 q−pr−q^2 +4s))/(256))u^2 −(((p^3 −4pq+8r)^2 )/(4096))=0 u=(√v) v^3 −((3p^2 −8q)/(16))v^2 +((3p^4 −16(p^2 q−pr−q^2 +4s))/(256))v−(((p^3 −4pq+8r)^2 )/(4096))=0 v=w+((3p^2 −8q)/(48)) w^3 +((3pr−q^2 −12s)/(48))w−((27p^2 s−9pqr+2q^3 −72qs+27r^3 )/(1728))=0 we need one real solution for w and then we go back w→v→u→γ α=−(p/4)−(1/(12))(√(3(48w+3p^2 −8q))) β=−w+(p^2 /8)−(q/3)+((p^3 −4pq+8r)/(8(√(3(48w+3p^2 −8q))))) γ=−(p/4)+(1/(12))(√(3(48w+3p^2 −8q))) δ=−w+(p^2 /8)−(q/3)−((p^3 −4pq+8r)/(8(√(3(48w+3p^2 −8q))))) if we found a “beautiful” w it′s fine, if not we can use a calculator to get good approximations

$$\mathrm{4}\:\mathrm{real}\:\mathrm{solutions}\:{x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} ,\:{x}_{\mathrm{3}} ,\:{x}_{\mathrm{4}} \:\mathrm{can}\:\mathrm{be}\:\mathrm{put}\:\mathrm{as}\:\mathrm{this}: \\ $$$${x}_{\mathrm{1}} =\alpha−\sqrt{\beta} \\ $$$${x}_{\mathrm{2}} =\alpha+\sqrt{\beta} \\ $$$${x}_{\mathrm{3}} =\gamma−\sqrt{\delta} \\ $$$${x}_{\mathrm{4}} =\gamma+\sqrt{\delta} \\ $$$$\mathrm{expanding}\:\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\left({x}−{x}_{\mathrm{3}} \right)\left({x}−{x}_{\mathrm{4}} \right)\:\mathrm{we}\:\mathrm{get} \\ $$$${x}^{\mathrm{4}} − \\ $$$$\:\:\:\:\:−\mathrm{2}\left(\alpha+\gamma\right){x}^{\mathrm{3}} + \\ $$$$\:\:\:\:\:+\left(\alpha^{\mathrm{2}} +\mathrm{4}\alpha\gamma−\beta+\gamma^{\mathrm{2}} −\delta\right){x}^{\mathrm{2}} − \\ $$$$\:\:\:\:\:−\mathrm{2}\left(\alpha^{\mathrm{2}} \gamma+\alpha\gamma^{\mathrm{2}} −\alpha\delta−\beta\gamma\right){x}+ \\ $$$$\:\:\:\:\:+\left(\alpha^{\mathrm{2}} −\beta\right)\left(\gamma^{\mathrm{2}} −\delta\right) \\ $$$$\mathrm{this}\:\mathrm{must}\:\mathrm{be}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as} \\ $$$${x}^{\mathrm{4}} +{px}^{\mathrm{3}} +{qx}^{\mathrm{2}} +{rx}+{s}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\left(\mathrm{1}\right)\:\:−\mathrm{2}\left(\alpha+\gamma\right)={p}}\\{\left(\mathrm{2}\right)\:\:\left(\alpha^{\mathrm{2}} +\mathrm{4}\alpha\gamma−\beta+\gamma^{\mathrm{2}} −\delta\right)={q}}\\{\left(\mathrm{3}\right)\:\:−\mathrm{2}\left(\alpha^{\mathrm{2}} \gamma+\alpha\gamma^{\mathrm{2}} −\alpha\delta−\beta\gamma\right)={r}}\\{\left(\mathrm{4}\right)\:\:\left(\alpha^{\mathrm{2}} −\beta\right)\left(\gamma^{\mathrm{2}} −\delta\right)={s}}\end{cases} \\ $$$$\mathrm{solve}\:\left(\mathrm{1}\right)\:\mathrm{for}\:\alpha,\:\left(\mathrm{2}\right)\:\mathrm{for}\:\beta,\:\left(\mathrm{3}\right)\:\mathrm{for}\:\delta,\:\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{4}\right) \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\gamma^{\mathrm{6}} +\frac{\mathrm{3}{p}}{\mathrm{2}}\gamma^{\mathrm{5}} +\frac{\mathrm{3}{p}^{\mathrm{2}} +\mathrm{2}{q}}{\mathrm{4}}\gamma^{\mathrm{4}} +\frac{{p}\left({p}^{\mathrm{2}} +\mathrm{4}{q}\right)}{\mathrm{8}}\gamma^{\mathrm{3}} +\frac{\mathrm{2}{p}^{\mathrm{2}} {q}+{pr}+{q}^{\mathrm{2}} −\mathrm{4}{s}}{\mathrm{16}}\gamma^{\mathrm{2}} +\frac{{p}\left({pr}+{q}^{\mathrm{2}} −\mathrm{4}{s}\right)}{\mathrm{32}}\gamma−\frac{{p}^{\mathrm{2}} {s}−{pqr}+{r}^{\mathrm{2}} }{\mathrm{64}}=\mathrm{0} \\ $$$$\gamma={u}−\frac{{p}}{\mathrm{4}} \\ $$$${u}^{\mathrm{6}} −\frac{\mathrm{3}{p}^{\mathrm{2}} −\mathrm{8}{q}}{\mathrm{16}}{u}^{\mathrm{4}} +\frac{\mathrm{3}{p}^{\mathrm{4}} −\mathrm{16}\left({p}^{\mathrm{2}} {q}−{pr}−{q}^{\mathrm{2}} +\mathrm{4}{s}\right)}{\mathrm{256}}{u}^{\mathrm{2}} −\frac{\left({p}^{\mathrm{3}} −\mathrm{4}{pq}+\mathrm{8}{r}\right)^{\mathrm{2}} }{\mathrm{4096}}=\mathrm{0} \\ $$$${u}=\sqrt{{v}} \\ $$$${v}^{\mathrm{3}} −\frac{\mathrm{3}{p}^{\mathrm{2}} −\mathrm{8}{q}}{\mathrm{16}}{v}^{\mathrm{2}} +\frac{\mathrm{3}{p}^{\mathrm{4}} −\mathrm{16}\left({p}^{\mathrm{2}} {q}−{pr}−{q}^{\mathrm{2}} +\mathrm{4}{s}\right)}{\mathrm{256}}{v}−\frac{\left({p}^{\mathrm{3}} −\mathrm{4}{pq}+\mathrm{8}{r}\right)^{\mathrm{2}} }{\mathrm{4096}}=\mathrm{0} \\ $$$${v}={w}+\frac{\mathrm{3}{p}^{\mathrm{2}} −\mathrm{8}{q}}{\mathrm{48}} \\ $$$${w}^{\mathrm{3}} +\frac{\mathrm{3}{pr}−{q}^{\mathrm{2}} −\mathrm{12}{s}}{\mathrm{48}}{w}−\frac{\mathrm{27}{p}^{\mathrm{2}} {s}−\mathrm{9}{pqr}+\mathrm{2}{q}^{\mathrm{3}} −\mathrm{72}{qs}+\mathrm{27}{r}^{\mathrm{3}} }{\mathrm{1728}}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{one}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{for}\:{w}\:\mathrm{and}\:\mathrm{then}\:\mathrm{we} \\ $$$$\mathrm{go}\:\mathrm{back}\:{w}\rightarrow{v}\rightarrow{u}\rightarrow\gamma \\ $$$$\alpha=−\frac{{p}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{12}}\sqrt{\mathrm{3}\left(\mathrm{48}{w}+\mathrm{3}{p}^{\mathrm{2}} −\mathrm{8}{q}\right)} \\ $$$$\beta=−{w}+\frac{{p}^{\mathrm{2}} }{\mathrm{8}}−\frac{{q}}{\mathrm{3}}+\frac{{p}^{\mathrm{3}} −\mathrm{4}{pq}+\mathrm{8}{r}}{\mathrm{8}\sqrt{\mathrm{3}\left(\mathrm{48}{w}+\mathrm{3}{p}^{\mathrm{2}} −\mathrm{8}{q}\right)}} \\ $$$$\gamma=−\frac{{p}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{12}}\sqrt{\mathrm{3}\left(\mathrm{48}{w}+\mathrm{3}{p}^{\mathrm{2}} −\mathrm{8}{q}\right)} \\ $$$$\delta=−{w}+\frac{{p}^{\mathrm{2}} }{\mathrm{8}}−\frac{{q}}{\mathrm{3}}−\frac{{p}^{\mathrm{3}} −\mathrm{4}{pq}+\mathrm{8}{r}}{\mathrm{8}\sqrt{\mathrm{3}\left(\mathrm{48}{w}+\mathrm{3}{p}^{\mathrm{2}} −\mathrm{8}{q}\right)}} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{found}\:\mathrm{a}\:``\mathrm{beautiful}''\:{w}\:\mathrm{it}'\mathrm{s}\:\mathrm{fine},\:\mathrm{if}\:\mathrm{not}\:\mathrm{we} \\ $$$$\mathrm{can}\:\mathrm{use}\:\mathrm{a}\:\mathrm{calculator}\:\mathrm{to}\:\mathrm{get}\:\mathrm{good}\:\mathrm{approximations} \\ $$

Commented by Tawa1 last updated on 03/Sep/18

Wow. God less you sir

$$\mathrm{Wow}.\:\mathrm{God}\:\mathrm{less}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by malwaan last updated on 04/Sep/18

Greet

$$\mathrm{Greet} \\ $$

Commented by MJS last updated on 05/Sep/18

in case of 2 real and 2 complex solutions put x_1 =α−(√β) x_2 =α+(√β) x_3 =γ−i(√δ) x_4 =γ+i(√δ) in case of 4 complex solutions put x_1 =α−(√β) x_2 =α+(√β) x_3 =γ−i(√δ) x_4 =γ+i(√δ) and follow the same procedure as above

$$\mathrm{in}\:\mathrm{case}\:\mathrm{of}\:\mathrm{2}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{put} \\ $$$${x}_{\mathrm{1}} =\alpha−\sqrt{\beta} \\ $$$${x}_{\mathrm{2}} =\alpha+\sqrt{\beta} \\ $$$${x}_{\mathrm{3}} =\gamma−\mathrm{i}\sqrt{\delta} \\ $$$${x}_{\mathrm{4}} =\gamma+\mathrm{i}\sqrt{\delta} \\ $$$$\mathrm{in}\:\mathrm{case}\:\mathrm{of}\:\mathrm{4}\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{put} \\ $$$${x}_{\mathrm{1}} =\alpha−\sqrt{\beta} \\ $$$${x}_{\mathrm{2}} =\alpha+\sqrt{\beta} \\ $$$${x}_{\mathrm{3}} =\gamma−\mathrm{i}\sqrt{\delta} \\ $$$${x}_{\mathrm{4}} =\gamma+\mathrm{i}\sqrt{\delta} \\ $$$$\mathrm{and}\:\mathrm{follow}\:\mathrm{the}\:\mathrm{same}\:\mathrm{procedure}\:\mathrm{as}\:\mathrm{above} \\ $$

Commented by Tawa1 last updated on 05/Sep/18

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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