Question Number 203002 by mr W last updated on 07/Jan/24
$${Solve}\:{for}\:{x}\in{R} \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{x}−\mathrm{3}=\mathrm{0} \\ $$
Answered by ajfour last updated on 07/Jan/24
$${p}^{\mathrm{2}} \left({p}^{\mathrm{4}} +\mathrm{12}\right)=\mathrm{16} \\ $$$$\Rightarrow\:\:\left({p}^{\mathrm{2}} /\mathrm{2}\right)^{\mathrm{3}} +\mathrm{3}\left({p}^{\mathrm{2}} /\mathrm{2}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\frac{{p}^{\mathrm{2}} }{\mathrm{2}}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$${now}\:{x}\:{is}\:{easy}. \\ $$
Commented by ajfour last updated on 07/Jan/24
Answered by mr W last updated on 07/Jan/24
![solve x^4 +4x−3=0 let′s generally look at the equation x^4 +px+q=0. say f(x)=x^4 +px+q lim_(x→−∞) f(x)=+∞ lim_(x→+∞) f(x)=+∞ f′(x)=4x^3 +p=0 ⇒x=−((p/4))^(1/3) that means f(x) has minimum at x_m =−((p/4))^(1/3) . f(x)_(min) =[−((p/4))^(1/3) ]^4 +p[−((p/4))^(1/3) ]+q=−3((p/4))^(4/3) +q there are three cases, see diagram below. case 1: f(x)_(min) =−3((p/4))^(4/3) +q>0 i.e. ((p/4))^4 −((q/3))^3 <0 ⇒f(x)=0 has no real root. case 2: f(x)_(min) =−3((p/4))^(4/3) +q=0 i.e. ((p/4))^4 −((q/3))^3 =0 ⇒f(x)=0 has a double real root x_(1,2) =x_m = −((p/4))^(1/3) case 3: f(x)_(min) =−3((p/4))^(4/3) +q=p<0 i.e. ((p/4))^4 −((q/3))^3 >0 ⇒f(x)=0 has two real roots.](https://www.tinkutara.com/question/Q203007.png)
$${solve}\:{x}^{\mathrm{4}} +\mathrm{4}{x}−\mathrm{3}=\mathrm{0} \\ $$$$ \\ $$$${let}'{s}\:{generally}\:{look}\:{at}\:{the}\:{equation} \\ $$$${x}^{\mathrm{4}} +{px}+{q}=\mathrm{0}. \\ $$$${say}\:{f}\left({x}\right)={x}^{\mathrm{4}} +{px}+{q} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right)=+\infty \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)=+\infty \\ $$$${f}'\left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} +{p}=\mathrm{0}\:\Rightarrow{x}=−\sqrt[{\mathrm{3}}]{\frac{{p}}{\mathrm{4}}} \\ $$$${that}\:{means}\:{f}\left({x}\right)\:{has}\:{minimum}\:{at} \\ $$$${x}_{{m}} =−\left(\frac{{p}}{\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} . \\ $$$${f}\left({x}\right)_{{min}} =\left[−\left(\frac{{p}}{\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right]^{\mathrm{4}} +{p}\left[−\left(\frac{{p}}{\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right]+{q}=−\mathrm{3}\left(\frac{{p}}{\mathrm{4}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} +{q} \\ $$$${there}\:{are}\:{three}\:{cases},\:{see}\:{diagram} \\ $$$${below}. \\ $$$$ \\ $$$${case}\:\mathrm{1}:\: \\ $$$${f}\left({x}\right)_{{min}} =−\mathrm{3}\left(\frac{{p}}{\mathrm{4}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} +{q}>\mathrm{0} \\ $$$${i}.{e}.\:\left(\frac{{p}}{\mathrm{4}}\right)^{\mathrm{4}} −\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} <\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{0}\:{has}\:{no}\:{real}\:{root}. \\ $$$$ \\ $$$${case}\:\mathrm{2}: \\ $$$${f}\left({x}\right)_{{min}} =−\mathrm{3}\left(\frac{{p}}{\mathrm{4}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} +{q}=\mathrm{0} \\ $$$${i}.{e}.\:\left(\frac{{p}}{\mathrm{4}}\right)^{\mathrm{4}} −\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{0}\:{has}\:{a}\:{double}\:{real}\:{root} \\ $$$${x}_{\mathrm{1},\mathrm{2}} ={x}_{{m}} =\:−\left(\frac{{p}}{\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$ \\ $$$${case}\:\mathrm{3}: \\ $$$${f}\left({x}\right)_{{min}} =−\mathrm{3}\left(\frac{{p}}{\mathrm{4}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} +{q}={p}<\mathrm{0} \\ $$$${i}.{e}.\:\left(\frac{{p}}{\mathrm{4}}\right)^{\mathrm{4}} −\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} >\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{0}\:{has}\:{two}\:{real}\:{roots}. \\ $$
Commented by mr W last updated on 07/Jan/24
Commented by mr W last updated on 07/Jan/24
![x^4 +px+q=0 (x^2 +ax+b)(x^2 +cx+d)=x^4 +px+q x^4 +(a+c)x^3 +(b+d+ac)x^2 +(bc+ad)x+bd=x^4 +px+q a+c=0 ⇒c=−a b+d+ac=0 ⇒d=a^2 −b=(a^2 /2)+(p/(2a)) bc+ad=p ⇒a^3 −2ba=p ⇒b=((a^3 −p)/(2a))=(a^2 /2)−(p/(2a)) bd=q ⇒b(a^2 −b)=q ⇒(((a^3 −p)/(2a)))(a^2 −((a^3 −p)/(2a)))=q ⇒a^6 −4qa^2 −p^2 =0 Δ=(−((4q)/3))^3 +(−(p^2 /2))^2 =64[((p/4))^4 −((q/3))^3 ] if ((p/4))^4 −((q/3))^3 ≥0, Δ≥0 ⇒a^2 =((8(√(((p/4))^4 −((q/3))^3 ))+(p^2 /2)))^(1/3) −((8(√(((p/4))^4 −((q/3))^3 ))−(p^2 /2)))^(1/3) ⇒a=±(√(((8(√(((p/4))^4 −((q/3))^3 ))+(p^2 /2)))^(1/3) −((8(√(((p/4))^4 −((q/3))^3 ))−(p^2 /2)))^(1/3) )) for real roots we take the same sign for a as the sign of p. example with p=4, q=−3 a=(√(2((((√2)+1))^(1/3) −(((√2)−1))^(1/3) ))) x_(1,2) =((−a±(√(a^2 −4b)))/2) =(1/2)(−a±(√(((2p)/a)−a^2 ))) ≈−1.7844 ∨ 0.6925](https://www.tinkutara.com/question/Q203010.png)
$${x}^{\mathrm{4}} +{px}+{q}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +{ax}+{b}\right)\left({x}^{\mathrm{2}} +{cx}+{d}\right)={x}^{\mathrm{4}} +{px}+{q} \\ $$$${x}^{\mathrm{4}} +\left({a}+{c}\right){x}^{\mathrm{3}} +\left({b}+{d}+{ac}\right){x}^{\mathrm{2}} +\left({bc}+{ad}\right){x}+{bd}={x}^{\mathrm{4}} +{px}+{q} \\ $$$${a}+{c}=\mathrm{0}\:\Rightarrow{c}=−{a} \\ $$$${b}+{d}+{ac}=\mathrm{0}\:\Rightarrow{d}={a}^{\mathrm{2}} −{b}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+\frac{{p}}{\mathrm{2}{a}} \\ $$$${bc}+{ad}={p}\:\Rightarrow{a}^{\mathrm{3}} −\mathrm{2}{ba}={p}\:\Rightarrow{b}=\frac{{a}^{\mathrm{3}} −{p}}{\mathrm{2}{a}}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−\frac{{p}}{\mathrm{2}{a}} \\ $$$${bd}={q}\:\Rightarrow{b}\left({a}^{\mathrm{2}} −{b}\right)={q}\: \\ $$$$\Rightarrow\left(\frac{{a}^{\mathrm{3}} −{p}}{\mathrm{2}{a}}\right)\left({a}^{\mathrm{2}} −\frac{{a}^{\mathrm{3}} −{p}}{\mathrm{2}{a}}\right)={q} \\ $$$$\Rightarrow{a}^{\mathrm{6}} −\mathrm{4}{qa}^{\mathrm{2}} −{p}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left(−\frac{\mathrm{4}{q}}{\mathrm{3}}\right)^{\mathrm{3}} +\left(−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{64}\left[\left(\frac{{p}}{\mathrm{4}}\right)^{\mathrm{4}} −\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} \right] \\ $$$${if}\:\left(\frac{{p}}{\mathrm{4}}\right)^{\mathrm{4}} −\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} \geqslant\mathrm{0},\:\Delta\geqslant\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\mathrm{8}\sqrt{\left(\frac{{p}}{\mathrm{4}}\right)^{\mathrm{4}} −\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\mathrm{8}\sqrt{\left(\frac{{p}}{\mathrm{4}}\right)^{\mathrm{4}} −\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\Rightarrow{a}=\pm\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{8}\sqrt{\left(\frac{{p}}{\mathrm{4}}\right)^{\mathrm{4}} −\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\mathrm{8}\sqrt{\left(\frac{{p}}{\mathrm{4}}\right)^{\mathrm{4}} −\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}}} \\ $$$${for}\:{real}\:{roots}\:{we}\:{take}\:{the}\:{same}\:{sign} \\ $$$${for}\:{a}\:{as}\:{the}\:{sign}\:{of}\:{p}. \\ $$$$ \\ $$$${example}\:{with}\:{p}=\mathrm{4},\:{q}=−\mathrm{3} \\ $$$${a}=\sqrt{\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}−\mathrm{1}}\right)} \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\frac{−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(−{a}\pm\sqrt{\frac{\mathrm{2}{p}}{{a}}−{a}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\approx−\mathrm{1}.\mathrm{7844}\:\vee\:\mathrm{0}.\mathrm{6925} \\ $$
Commented by ajfour last updated on 07/Jan/24
yes this is the simpler in fact. Thank you sir.
but b=0 in my formula for x would be equivalently simple too.
Commented by mr W last updated on 07/Jan/24
$${you}\:{are}\:{right}\:{sir}!\:{thanks}! \\ $$
Answered by Frix last updated on 07/Jan/24
$$\mathrm{Just}\:\mathrm{the}\:\mathrm{same}\:\mathrm{but}\:\mathrm{another}\:\mathrm{point}\:\mathrm{of}\:\mathrm{view}. \\ $$$${x}^{\mathrm{4}} +{px}+{q}=\mathrm{0} \\ $$$${p}=−\mathrm{4}\alpha\left(\alpha^{\mathrm{2}} +\beta\right) \\ $$$${q}=\left(\alpha^{\mathrm{2}} −\beta\right)\left(\mathrm{3}\alpha^{\mathrm{2}} +\beta\right) \\ $$$${x}^{\mathrm{4}} −\mathrm{4}\alpha\left(\alpha^{\mathrm{2}} +\beta\right){x}+\left(\alpha^{\mathrm{2}} −\beta\right)\left(\mathrm{3}\alpha^{\mathrm{2}} +\beta\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}\alpha{x}+\alpha^{\mathrm{2}} −\beta\right)\left({x}^{\mathrm{2}} +\mathrm{2}\alpha{x}+\mathrm{3}\alpha^{\mathrm{2}} −\beta\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1},\:\mathrm{2}} =\alpha\pm\sqrt{\beta} \\ $$$${x}_{\mathrm{2},\:\mathrm{3}} =−\alpha\pm\sqrt{−\left(\mathrm{2}\alpha^{\mathrm{2}} +\beta\right)} \\ $$$$\mathrm{We}\:\mathrm{end}\:\mathrm{up}\:\mathrm{with} \\ $$$$\alpha^{\mathrm{6}} −\frac{{q}\alpha^{\mathrm{2}} }{\mathrm{4}}−\frac{{p}^{\mathrm{2}} }{\mathrm{64}}=\mathrm{0}\wedge\beta=−\alpha^{\mathrm{2}} −\frac{{p}}{\mathrm{4}\alpha} \\ $$
Commented by mr W last updated on 07/Jan/24
$${thanks}\:{sir}! \\ $$