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-

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

solving

{a x}^{4} + {b x}^{3} + {c x}^{2} + d x + e = 0

(a \neq 0, b, c, d, e) \in Q

special cases (easy to solve)

{a x}^{4} + e = 0 solve {a t}^{2} + e = 0 \Rightarrow x = \pm \sqrt{t_{1, 2}}

{a x}^{4} + {c x}^{2} + e = 0 solve {a t}^{2} + c t + e = 0 \Rightarrow x = \pm \sqrt{t_{1, 2}}

always try all factors of \pm e

because a (x - α) (x - β) (x - γ) (x - δ) = {a x}^{4} + \dots + α β γ δ

\Rightarrow e = α β γ δ

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 \in Q \Rightarrow 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} + {p x}^{3} + {q x}^{2} + r x + s = 0

[p = \frac{b}{a} q = \frac{c}{a} r = \frac{d}{a} s = \frac{e}{a}]

I^{'} ll soon post some cases I^{'} ve been able to solve

as comments

Commented by Tawa1 last updated on 03/Sep/18

Great sir . God bless you 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

4 real solutions x_{1}, x_{2}, x_{3}, x_{4} can be put as this :

x_{1} = α - \sqrt{β}

x_{2} = α + \sqrt{β}

x_{3} = γ - \sqrt{δ}

x_{4} = γ + \sqrt{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} + 4 α γ - β + γ^{2} - δ) x^{2} -

- 2 (α^{2} γ + α γ^{2} - α δ - β γ) x +

+ (α^{2} - β) (γ^{2} - δ)

this must be the same as

x^{4} + {p x}^{3} + {q x}^{2} + r x + s \Rightarrow

\Rightarrow {\begin{cases} (1) - 2 (α + γ) = p \\ (2) (α^{2} + 4 α γ - β + γ^{2} - δ) = q \\ (3) - 2 (α^{2} γ + α γ^{2} - α δ - β γ) = r \\ (4) (α^{2} - β) (γ^{2} - δ) = s \end{cases}

solve (1) for α, (2) for β, (3) for δ, insert in (4)

which leads to

γ^{6} + \frac{3 p}{2} γ^{5} + \frac{3 p^{2} + 2 q}{4} γ^{4} + \frac{p (p^{2} + 4 q)}{8} γ^{3} + \frac{2 p^{2} q + p r + q^{2} - 4 s}{16} γ^{2} + \frac{p (p r + q^{2} - 4 s)}{32} γ - \frac{p^{2} s - p q r + r^{2}}{64} = 0

γ = u - \frac{p}{4}

u^{6} - \frac{3 p^{2} - 8 q}{16} u^{4} + \frac{3 p^{4} - 16 (p^{2} q - p r - q^{2} + 4 s)}{256} u^{2} - \frac{{(p^{3} - 4 p q + 8 r)}^{2}}{4096} = 0

u = \sqrt{v}

v^{3} - \frac{3 p^{2} - 8 q}{16} v^{2} + \frac{3 p^{4} - 16 (p^{2} q - p r - q^{2} + 4 s)}{256} v - \frac{{(p^{3} - 4 p q + 8 r)}^{2}}{4096} = 0

v = w + \frac{3 p^{2} - 8 q}{48}

w^{3} + \frac{3 p r - q^{2} - 12 s}{48} w - \frac{27 p^{2} s - 9 p q r + 2 q^{3} - 72 q s + 27 r^{3}}{1728} = 0

we need one real solution for w and then we

go back w \to v \to u \to γ

α = - \frac{p}{4} - \frac{1}{12} \sqrt{3 (48 w + 3 p^{2} - 8 q)}

β = - w + \frac{p^{2}}{8} - \frac{q}{3} + \frac{p^{3} - 4 p q + 8 r}{8 \sqrt{3 (48 w + 3 p^{2} - 8 q)}}

γ = - \frac{p}{4} + \frac{1}{12} \sqrt{3 (48 w + 3 p^{2} - 8 q)}

δ = - w + \frac{p^{2}}{8} - \frac{q}{3} - \frac{p^{3} - 4 p q + 8 r}{8 \sqrt{3 (48 w + 3 p^{2} - 8 q)}}

if we found a “ {beautiful}^{″} w {it}^{'} s fine, if not we

can use a calculator to get good approximations

Commented by Tawa1 last updated on 03/Sep/18

Wow . God less you sir

Commented by malwaan last updated on 04/Sep/18

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

in case of 2 real and 2 complex solutions put

x_{1} = α - \sqrt{β}

x_{2} = α + \sqrt{β}

x_{3} = γ - i \sqrt{δ}

x_{4} = γ + i \sqrt{δ}

in case of 4 complex solutions put

x_{1} = α - \sqrt{β}

x_{2} = α + \sqrt{β}

x_{3} = γ - i \sqrt{δ}

x_{4} = γ + i \sqrt{δ}

and follow the same procedure as above

Commented by Tawa1 last updated on 05/Sep/18

God bless you sir .

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