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


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

$solving$ ${a x}^{4} + {b x}^{3} + {c x}^{2} + d x + e = 0$ $(a \neq 0, b, c, d, e) \in Q$ $special cases (easy to solve)$ ${a x}^{4} + e = 0 solve {a t}^{2} + e = 0 \Rightarrow x = \pm \sqrt{t_{1, 2}}$ ${a x}^{4} + {c x}^{2} + e = 0 solve {a t}^{2} + c t + e = 0 \Rightarrow x = \pm \sqrt{t_{1, 2}}$ $always try all factors of \pm e$ $because a (x - α) (x - β) (x - γ) (x - δ) = {a x}^{4} + . . . + α β γ δ$ $\Rightarrow e = α β γ δ$ $next we must find the nature of the solutions$ $4 real solutions$ $2 real & 2 complex solutions$ $4 complex solutions$ $a, b, c, d, e \in Q \Rightarrow complex solutions always in$ $conjugated pairs$ $draw the function or calculate some values$ $to find the number of real solutions$ $divide by a$ $x^{4} + {p x}^{3} + {q x}^{2} + r x + s = 0$ $[p = \frac{b}{a} q = \frac{c}{a} r = \frac{d}{a} s = \frac{e}{a}]$ $I^{'} ll soon post some cases I^{'} ve been able to solve$ $as comments$
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

$in case of 2 real and 2 complex solutions put$ $x_{1} = α - \sqrt{β}$ $x_{2} = α + \sqrt{β}$ $x_{3} = γ - i \sqrt{δ}$ $x_{4} = γ + i \sqrt{δ}$ $in case of 4 complex solutions put$ $x_{1} = α - \sqrt{β}$ $x_{2} = α + \sqrt{β}$ $x_{3} = γ - i \sqrt{δ}$ $x_{4} = γ + i \sqrt{δ}$ $and follow the same procedure as above$
Commented by Tawa1 last updated on 05/Sep/18

$God bless you sir .$
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