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Question Number 156404 by ajfour last updated on 10/Oct/21
x^4 +bx^2 +cx+d=0 let cx=m+px^2 +x^4 ⇒ 2x^4 +(b+p)x^2 +m+d=0 x^2 =−(((b+p)/4))±(√((((b+p)/4))^2 −(((m+d)/2)))) (p−b)x^2 −2cx+m−d=0 x=(c/(p−b))±(√(((c/(p−b)))^2 −(((m−d)/(p−b))))) ⇒ (((b^2 −p^2 ))/4)±(√(((b^2 −p^2 )^2 −8(p−b)^2 (m+d))/(16))) +((2c^2 )/(b−p))∓(√((4c^4 −4c^2 (p−b)(m−d))/((p−b)^2 ))) +m−d=0 Just if m=d ⇒ (((b^2 −p^2 ))/4)±(√(((b^2 −p^2 )^2 −16d(p−b))/(16))) +((4c^2 )/(b−p))=0 ⇒ (((b^2 −p^2 )/4)+((4c^2 )/(b−p)))^2 +d(p−b)=(((b^2 −p^2 )^2 )/(16)) ⇒ ((16c^4 )/((b−p)^2 ))+(d+2c^2 )(b+p)=0 let b−p=z z^2 (2b−z)=−((16c^4 )/(2c^2 +d)) z^3 −2bz^2 −k=0 (k>0) if b=−1 , d=0, c→−c z^3 +2z^2 −8c^2 =0 let z=((2c)/t) 8c^3 +8c^2 t−8c^2 t^3 =0 ⇒ t^3 −t=c back to square one.
$$\:\:\mathrm{x}^{\mathrm{4}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{cx}+\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{cx}=\mathrm{m}+\mathrm{px}^{\mathrm{2}} +\mathrm{x}^{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{2x}^{\mathrm{4}} +\left(\mathrm{b}+\mathrm{p}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{m}+\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} =−\left(\frac{\mathrm{b}+\mathrm{p}}{\mathrm{4}}\right)\pm\sqrt{\left(\frac{\mathrm{b}+\mathrm{p}}{\mathrm{4}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{m}+\mathrm{d}}{\mathrm{2}}\right)} \\ $$$$\left(\mathrm{p}−\mathrm{b}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{2cx}+\mathrm{m}−\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{c}}{\mathrm{p}−\mathrm{b}}\pm\sqrt{\left(\frac{\mathrm{c}}{\mathrm{p}−\mathrm{b}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{m}−\mathrm{d}}{\mathrm{p}−\mathrm{b}}\right)} \\ $$$$\Rightarrow \\ $$$$\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)}{\mathrm{4}}\pm\sqrt{\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{p}−\mathrm{b}\right)^{\mathrm{2}} \left(\mathrm{m}+\mathrm{d}\right)}{\mathrm{16}}} \\ $$$$+\frac{\mathrm{2c}^{\mathrm{2}} }{\mathrm{b}−\mathrm{p}}\mp\sqrt{\frac{\mathrm{4c}^{\mathrm{4}} −\mathrm{4c}^{\mathrm{2}} \left(\mathrm{p}−\mathrm{b}\right)\left(\mathrm{m}−\mathrm{d}\right)}{\left(\mathrm{p}−\mathrm{b}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:+\mathrm{m}−\mathrm{d}=\mathrm{0} \\ $$$${Just}\:{if}\:\:{m}={d}\:\:\Rightarrow \\ $$$$\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)}{\mathrm{4}}\pm\sqrt{\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{16d}\left(\mathrm{p}−\mathrm{b}\right)}{\mathrm{16}}} \\ $$$$+\frac{\mathrm{4c}^{\mathrm{2}} }{\mathrm{b}−\mathrm{p}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{4c}^{\mathrm{2}} }{\mathrm{b}−\mathrm{p}}\right)^{\mathrm{2}} +\mathrm{d}\left(\mathrm{p}−\mathrm{b}\right)=\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\Rightarrow\:\frac{\mathrm{16c}^{\mathrm{4}} }{\left(\mathrm{b}−\mathrm{p}\right)^{\mathrm{2}} }+\left(\mathrm{d}+\mathrm{2c}^{\mathrm{2}} \right)\left(\mathrm{b}+\mathrm{p}\right)=\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{b}−\mathrm{p}=\mathrm{z} \\ $$$$\mathrm{z}^{\mathrm{2}} \left(\mathrm{2b}−\mathrm{z}\right)=−\frac{\mathrm{16c}^{\mathrm{4}} }{\mathrm{2c}^{\mathrm{2}} +\mathrm{d}} \\ $$$$\mathrm{z}^{\mathrm{3}} −\mathrm{2bz}^{\mathrm{2}} −\mathrm{k}=\mathrm{0}\:\:\:\:\left(\mathrm{k}>\mathrm{0}\right) \\ $$$$\mathrm{if}\:\mathrm{b}=−\mathrm{1}\:,\:\mathrm{d}=\mathrm{0},\:\mathrm{c}\rightarrow−\mathrm{c} \\ $$$$\mathrm{z}^{\mathrm{3}} +\mathrm{2z}^{\mathrm{2}} −\mathrm{8c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{z}=\frac{\mathrm{2c}}{\mathrm{t}} \\ $$$$\mathrm{8c}^{\mathrm{3}} +\mathrm{8c}^{\mathrm{2}} \mathrm{t}−\mathrm{8c}^{\mathrm{2}} \mathrm{t}^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{t}^{\mathrm{3}} −\mathrm{t}=\mathrm{c} \\ $$$$\mathrm{back}\:\mathrm{to}\:\mathrm{square}\:\mathrm{one}. \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa11 last updated on 10/Oct/21
Sir, I thought you got one formula that work for power 4 already. You want to get another one?
$$\mathrm{Sir},\:\mathrm{I}\:\mathrm{thought}\:\mathrm{you}\:\mathrm{got}\:\mathrm{one}\:\mathrm{formula}\:\mathrm{that}\:\mathrm{work}\:\mathrm{for}\:\mathrm{power}\:\mathrm{4}\:\mathrm{already}. \\ $$$$\mathrm{You}\:\mathrm{want}\:\mathrm{to}\:\mathrm{get}\:\mathrm{another}\:\mathrm{one}? \\ $$
Commented by ajfour last updated on 11/Oct/21
I havent yet derived a very genersl one that doesnt need to adopt imaginary numbers and inverse trigonometry.
$$\mathrm{I}\:\mathrm{havent}\:\mathrm{yet}\:\mathrm{derived}\:\mathrm{a}\:\mathrm{very} \\ $$$$\mathrm{genersl}\:\mathrm{one}\:\mathrm{that}\:\mathrm{doesnt}\:\mathrm{need} \\ $$$$\mathrm{to}\:\mathrm{adopt}\:\mathrm{imaginary}\:\mathrm{numbers} \\ $$$$\mathrm{and}\:\mathrm{inverse}\:\mathrm{trigonometry}. \\ $$
Commented by Tawa11 last updated on 11/Oct/21
Ohh. Altight sir. God will help you more.
$$\mathrm{Ohh}.\:\mathrm{Altight}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{will}\:\mathrm{help}\:\mathrm{you}\:\mathrm{more}. \\ $$

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