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Question Number 123409 by Snail last updated on 25/Nov/20
x^4 +12x−5=0 show that the sum of two roots of the equation is 2
$${x}^{\mathrm{4}} +\mathrm{12}{x}−\mathrm{5}=\mathrm{0}\:\:\:{show}\:{that}\:{the}\:{sum}\:{of}\:{two}\:{roots} \\ $$$${of}\:{the}\:{equation}\:{is}\:\mathrm{2} \\ $$
Commented by ajfour last updated on 12/Dec/23
x=1±(√2) , x=−1±2i
$${x}=\mathrm{1}\pm\sqrt{\mathrm{2}}\:,\:\:{x}=−\mathrm{1}\pm\mathrm{2}{i} \\ $$
Answered by MJS_new last updated on 25/Nov/20
x^4 +12x−5=(x^2 −2x+5)(x^2 +2x−1) (x−a)(x−b)(x−α)(x−β)= =(x^2 −(a+b)x+ab)(x^2 −(α+β)x+αβ) (x^2 −(a+b)x+ab)=(x^2 −2x+5) ⇒ a+b=2 (x^2 −(α+β)x+αβ)=(x^2 +2x−1) ⇒ α+β=−2
$${x}^{\mathrm{4}} +\mathrm{12}{x}−\mathrm{5}=\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}\right) \\ $$$$ \\ $$$$\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−\alpha\right)\left({x}−\beta\right)= \\ $$$$=\left({x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}\right)\left({x}^{\mathrm{2}} −\left(\alpha+\beta\right){x}+\alpha\beta\right) \\ $$$$\left({x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}\right)=\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}\right)\:\Rightarrow\:{a}+{b}=\mathrm{2} \\ $$$$\left({x}^{\mathrm{2}} −\left(\alpha+\beta\right){x}+\alpha\beta\right)=\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}\right)\:\Rightarrow\:\alpha+\beta=−\mathrm{2} \\ $$
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
$$\mathrm{the}\:\mathrm{method}\:\mathrm{sometimes}\:\mathrm{gives}\:\mathrm{exact}\:\mathrm{and}\:“\mathrm{nice}'' \\ $$$$\mathrm{solutions}\:\mathrm{for} \\ $$$${ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{divide}\:\mathrm{by}\:{a} \\ $$$${x}^{\mathrm{4}} +\frac{{b}}{{a}}{x}^{\mathrm{3}} +\frac{{c}}{{a}}{x}^{\mathrm{2}} +\frac{{d}}{{a}}{x}+\frac{{e}}{{a}}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{rename}\:\mathrm{variables} \\ $$$${x}^{\mathrm{4}} +{hx}^{\mathrm{3}} +{jx}^{\mathrm{2}} +{kx}+{l}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{let}\:{x}={y}−\frac{{h}}{\mathrm{4}} \\ $$$${y}^{\mathrm{4}} −\frac{\mathrm{3}{h}^{\mathrm{2}} −\mathrm{8}{j}}{\mathrm{8}}{y}^{\mathrm{2}} +\frac{{h}^{\mathrm{3}} −\mathrm{4}{hj}+\mathrm{8}{k}}{\mathrm{8}}{y}+\frac{{h}^{\mathrm{2}} {j}−\mathrm{4}{hk}+\mathrm{26}{l}}{\mathrm{16}}=\mathrm{0} \\ $$$$\mathrm{rename}\:\mathrm{again} \\ $$$${y}^{\mathrm{4}} +{py}^{\mathrm{2}} +{qy}+{r}=\mathrm{0} \\ $$$$\mathrm{now}\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{2}\:\mathrm{square}\:\mathrm{factors} \\ $$$${y}^{\mathrm{4}} +{py}^{\mathrm{2}} +{qy}+{r}=\left({y}^{\mathrm{2}} −\alpha{y}−\beta\right)\left({y}^{\mathrm{2}} +\alpha{y}−\gamma\right) \\ $$$$\Leftrightarrow \\ $$$$\begin{cases}{{p}=−\alpha^{\mathrm{2}} −\beta−\gamma}\\{{q}=−\alpha\left(\beta−\gamma\right)}\\{{r}=\beta\gamma}\end{cases} \\ $$$$\mathrm{now}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{for}\:\gamma\:\mathrm{and}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{for}\:\beta \\ $$$$\mathrm{and}\:\mathrm{insert}\:\mathrm{both}\:\mathrm{into}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \\ $$$$\Rightarrow \\ $$$$\begin{cases}{\gamma=−\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}−\frac{{p}}{\mathrm{2}}+\frac{{q}}{\mathrm{2}\alpha}}\\{\beta=−\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}−\frac{{p}}{\mathrm{2}}+\frac{{q}}{\mathrm{2}\alpha}}\\{{r}=\frac{\alpha^{\mathrm{4}} }{\mathrm{4}}+\frac{{p}\alpha^{\mathrm{2}} }{\mathrm{2}}+\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−\frac{{q}^{\mathrm{2}} }{\mathrm{4}\alpha^{\mathrm{2}} }=\mathrm{0}}\end{cases} \\ $$$$\mathrm{transform}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \\ $$$$\alpha^{\mathrm{6}} +\mathrm{2}{p}\alpha^{\mathrm{4}} +\left({p}^{\mathrm{2}} −\mathrm{4}{r}\right)\alpha^{\mathrm{2}} −{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{let}\:\alpha=\sqrt{{t}} \\ $$$${t}^{\mathrm{3}} +\mathrm{2}{pt}^{\mathrm{2}} +\left({p}^{\mathrm{2}} −\mathrm{4}{r}\right){t}−{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{let}\:{t}={z}−\frac{\mathrm{2}{p}}{\mathrm{3}} \\ $$$${z}^{\mathrm{3}} −\frac{{p}^{\mathrm{2}} +\mathrm{12}{r}}{\mathrm{3}}{z}−\frac{\mathrm{2}{p}^{\mathrm{3}} +\mathrm{72}{pr}+\mathrm{27}{q}^{\mathrm{2}} }{\mathrm{27}}=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{this}\:\mathrm{has}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:“\mathrm{nice}''\:\mathrm{solution}\:\in\mathbb{R}, \\ $$$$\mathrm{that}'\mathrm{s}\:\mathrm{fine};\:\mathrm{if}\:\mathrm{not},\:\mathrm{better}\:\mathrm{solve}\:\mathrm{approximately} \\ $$$$\mathrm{remember}\:\mathrm{we}\:\mathrm{must}\:\mathrm{go}\:\mathrm{back}\:\mathrm{to}\:\alpha=\sqrt{{z}−\frac{\mathrm{2}{p}}{\mathrm{3}}} \\ $$$$\mathrm{so}\:{z}\:\mathrm{being}\:{not}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{number}\:\mathrm{will}\:\mathrm{make} \\ $$$$\mathrm{things}\:\mathrm{worse} \\ $$
Commented by Snail last updated on 25/Nov/20
How do u know that the biquadratic will be factorized in (x^2 −2x+5)(x^2 +2x−1)...
$$ \\ $$$${How}\:{do}\:{u}\:{know}\:{that}\:{the}\:{biquadratic}\:{will}\:{be}\: \\ $$$${factorized}\:{in}\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}\right)… \\ $$
Commented by MJS_new last updated on 25/Nov/20
I will post the method. just give me a little time
$$\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{the}\:\mathrm{method}.\:\mathrm{just}\:\mathrm{give}\:\mathrm{me}\:\mathrm{a}\:\mathrm{little} \\ $$$$\mathrm{time} \\ $$

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