x^4 +12x−5=0 show that the sum of two roots of the equation is 2


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Question Number 123409 by Snail last updated on 25/Nov/20

$x^{4} + 12 x - 5 = 0 s h o w t h a t t h e s u m o f t w o r o o t s$ $o f t h e e q u a t i o n i s 2$
	Answered by MJS_new last updated on 25/Nov/20

	$x^{4} + 12 x - 5 = (x^{2} - 2 x + 5) (x^{2} + 2 x - 1)$ $(x - a) (x - b) (x - α) (x - β) =$ $= (x^{2} - (a + b) x + a b) (x^{2} - (α + β) x + α β)$ $(x^{2} - (a + b) x + a b) = (x^{2} - 2 x + 5) \Rightarrow a + b = 2$ $(x^{2} - (α + β) x + α β) = (x^{2} + 2 x - 1) \Rightarrow α + β = - 2$
	Commented by MJS_new last updated on 25/Nov/20
	$the method sometimes gives exact and “nice” solutions for ax^4 +bx^3 +cx^2 +dx+e=0 (1) divide by a x^4 +(b/a)x^3 +(c/a)x^2 +(d/a)x+(e/a)=0 and rename variables x^4 +hx^3 +jx^2 +kx+l=0 (2) let x=y−(h/4) y^4 −((3h^2 −8j)/8)y^2 +((h^3 −4hj+8k)/8)y+((h^2 j−4hk+26l)/(16))=0 rename again y^4 +py^2 +qy+r=0 now try to find 2 square factors y^4 +py^2 +qy+r=(y^2 −αy−β)(y^2 +αy−γ) ⇔ { ((p=−α^2 −β−γ)),((q=−α(β−γ))),((r=βγ)) :} now solve the 1^(st) for γ and the 2^(nd) for β and insert both into the 3^(rd) ⇒ { ((γ=−(α^2 /2)−(p/2)+(q/(2α)))),((β=−(α^2 /2)−(p/2)+(q/(2α)))),((r=(α^4 /4)+((pα^2 )/2)+(p^2 /4)−(q^2 /(4α^2 ))=0)) :} transform the 3^(rd) α^6 +2pα^4 +(p^2 −4r)α^2 −q^2 =0 let α=(√t) t^3 +2pt^2 +(p^2 −4r)t−q^2 =0 let t=z−((2p)/3) z^3 −((p^2 +12r)/3)z−((2p^3 +72pr+27q^2 )/(27))=0 if this has at least one “nice” solution ∈R, that′s fine; if not, better solve approximately remember we must go back to α=(√(z−((2p)/3))) so z being not a simple number will make things worse$
	$the method sometimes gives exact and ‘ ‘ {nice}^{″}$ $solutions for$ ${a x}^{4} + {b x}^{3} + {c x}^{2} + d x + e = 0$ $(1) divide by a$ $x^{4} + \frac{b}{a} x^{3} + \frac{c}{a} x^{2} + \frac{d}{a} x + \frac{e}{a} = 0$ $and rename variables$ $x^{4} + {h x}^{3} + {j x}^{2} + k x + l = 0$ $(2) let x = y - \frac{h}{4}$ $y^{4} - \frac{3 h^{2} - 8 j}{8} y^{2} + \frac{h^{3} - 4 h j + 8 k}{8} y + \frac{h^{2} j - 4 h k + 26 l}{16} = 0$ $rename again$ $y^{4} + {p y}^{2} + q y + r = 0$ $now try to find 2 square factors$ $y^{4} + {p y}^{2} + q y + r = (y^{2} - α y - β) (y^{2} + α y - γ)$ $\Leftrightarrow$ ${\begin{cases} p = - α^{2} - β - γ \\ q = - α (β - γ) \\ r = β γ \end{cases}$ $now solve the 1^{st} for γ and the 2^{nd} for β$ $and insert both into the 3^{rd}$ $\Rightarrow$ ${\begin{cases} γ = - \frac{α^{2}}{2} - \frac{p}{2} + \frac{q}{2 α} \\ β = - \frac{α^{2}}{2} - \frac{p}{2} + \frac{q}{2 α} \\ r = \frac{α^{4}}{4} + \frac{p α^{2}}{2} + \frac{p^{2}}{4} - \frac{q^{2}}{4 α^{2}} = 0 \end{cases}$ $transform the 3^{rd}$ $α^{6} + 2 p α^{4} + (p^{2} - 4 r) α^{2} - q^{2} = 0$ $let α = \sqrt{t}$ $t^{3} + 2 {p t}^{2} + (p^{2} - 4 r) t - q^{2} = 0$ $let t = z - \frac{2 p}{3}$ $z^{3} - \frac{p^{2} + 12 r}{3} z - \frac{2 p^{3} + 72 p r + 27 q^{2}}{27} = 0$ $if this has at least one ‘ ‘ {nice}^{″} solution \in R,$ ${that}^{'} s fine; if not, better solve approximately$ $remember we must go back to α = \sqrt{z - \frac{2 p}{3}}$ $so z being n o t a simple number will make$ $things worse$
	Commented by Snail last updated on 25/Nov/20

	$H o w d o u k n o w t h a t t h e b i q u a d r a t i c w i l l b e$ $f a c t o r i z e d i n (x^{2} - 2 x + 5) (x^{2} + 2 x - 1) . . .$
	Commented by MJS_new last updated on 25/Nov/20

	$I will post the method . just give me a little$ $time$
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