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Question Number 26235 by ajfour last updated on 22/Dec/17

Find the real root of x^2 +(1/x)=c .

$${Find}\:{the}\:{real}\:{root}\:{of} \\ $$$$\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}={c}\:. \\ $$

Answered by mrW1 last updated on 22/Dec/17

x≠0 x^3 −cx+1=0 Δ=((1/2))^2 +(−(c/3))^3 =(1/4)−(c^3 /(27)) Δ=0⇒c=(3/(^3 (√4))) case 1: if c=(3/(^3 (√4))), two real solutions. x_1 =2 ^3 (√(−(1/2)))=−(2/(^3 (√2)))≈−1.587 x_2 =− ^3 (√(−(1/2)))=(1/(^3 (√2)))≈0.794 case 2: if c<(3/(^3 (√4))), one real solution. x=^3 (√(−(1/2)+(√Δ)))−^3 (√((1/2)+(√Δ))) e.g. c=1 ⇒ Δ=(1/4)−(1/(27))=((23)/(108)) x=^3 (√(−(1/2)+(√((23)/(108)))))−^3 (√((1/2)+(√((23)/(108)))))≈−1.324 case 3: if c>(3/(^3 (√4))), three real solutions. e.g. c=3, r=(√(c^3 /(27)))=1 cos ∅=−(1/(2r))=−(1/2) ⇒∅=((2π)/3) x_1 =2 ^3 (√r) cos (∅/3)=2 cos ((2π)/9)≈1.532 x_2 =2 ^3 (√r) cos ((∅+2π)/3)=2 cos ((8π)/9)≈−1.879 x_3 =2 ^3 (√r) cos ((∅+4π)/3)=2 cos ((14π)/9)≈0.347

$${x}\neq\mathrm{0} \\ $$$${x}^{\mathrm{3}} −{cx}+\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(−\frac{{c}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{{c}^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\Delta=\mathrm{0}\Rightarrow{c}=\frac{\mathrm{3}}{\:^{\mathrm{3}} \sqrt{\mathrm{4}}} \\ $$$${case}\:\mathrm{1}:\:{if}\:{c}=\frac{\mathrm{3}}{\:^{\mathrm{3}} \sqrt{\mathrm{4}}},\:{two}\:{real}\:{solutions}. \\ $$$${x}_{\mathrm{1}} =\mathrm{2}\:\:^{\mathrm{3}} \sqrt{−\frac{\mathrm{1}}{\mathrm{2}}}=−\frac{\mathrm{2}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\approx−\mathrm{1}.\mathrm{587} \\ $$$${x}_{\mathrm{2}} =−\:\:^{\mathrm{3}} \sqrt{−\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\approx\mathrm{0}.\mathrm{794} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{if}\:{c}<\frac{\mathrm{3}}{\:^{\mathrm{3}} \sqrt{\mathrm{4}}},\:{one}\:{real}\:{solution}. \\ $$$${x}=\:^{\mathrm{3}} \sqrt{−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\Delta}}−\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\Delta}} \\ $$$${e}.{g}.\:{c}=\mathrm{1}\:\Rightarrow\:\Delta=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{27}}=\frac{\mathrm{23}}{\mathrm{108}} \\ $$$${x}=\:^{\mathrm{3}} \sqrt{−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{23}}{\mathrm{108}}}}−\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{23}}{\mathrm{108}}}}\approx−\mathrm{1}.\mathrm{324} \\ $$$$ \\ $$$${case}\:\mathrm{3}:\:{if}\:{c}>\frac{\mathrm{3}}{\:^{\mathrm{3}} \sqrt{\mathrm{4}}},\:{three}\:{real}\:{solutions}. \\ $$$${e}.{g}.\:{c}=\mathrm{3}, \\ $$$${r}=\sqrt{\frac{{c}^{\mathrm{3}} }{\mathrm{27}}}=\mathrm{1} \\ $$$$\mathrm{cos}\:\emptyset=−\frac{\mathrm{1}}{\mathrm{2}{r}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\emptyset=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${x}_{\mathrm{1}} =\mathrm{2}\:\:^{\mathrm{3}} \sqrt{{r}}\:\mathrm{cos}\:\frac{\emptyset}{\mathrm{3}}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\approx\mathrm{1}.\mathrm{532} \\ $$$${x}_{\mathrm{2}} =\mathrm{2}\:\:^{\mathrm{3}} \sqrt{{r}}\:\mathrm{cos}\:\frac{\emptyset+\mathrm{2}\pi}{\mathrm{3}}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{9}}\approx−\mathrm{1}.\mathrm{879} \\ $$$${x}_{\mathrm{3}} =\mathrm{2}\:\:^{\mathrm{3}} \sqrt{{r}}\:\mathrm{cos}\:\frac{\emptyset+\mathrm{4}\pi}{\mathrm{3}}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{14}\pi}{\mathrm{9}}\approx\mathrm{0}.\mathrm{347} \\ $$

Commented by ajfour last updated on 23/Dec/17

please sir, i dont follow this. Help me derive formula for this. how do you obtain Δ as a function of c ? please see Q.26251 some consrruction method should be there. Moreover all cubic equations can be reduced to this cubic equation, i think..

$${please}\:{sir},\:{i}\:{dont}\:{follow}\:{this}. \\ $$$${Help}\:{me}\:{derive}\:{formula}\:{for}\:{this}. \\ $$$${how}\:{do}\:{you}\:{obtain}\:\Delta\:{as}\:{a}\:{function} \\ $$$${of}\:{c}\:?\:{please}\:{see}\:{Q}.\mathrm{26251} \\ $$$${some}\:{consrruction}\:{method}\: \\ $$$${should}\:{be}\:{there}.\:{Moreover}\:{all} \\ $$$${cubic}\:{equations}\:{can}\:{be}\:{reduced} \\ $$$${to}\:{this}\:{cubic}\:{equation},\:{i}\:{think}.. \\ $$

Commented by mrW1 last updated on 23/Dec/17

Yes, every cubic equation can be reduced to the form x^3 +px+q=0 ...(i) let v=(p/(3u)), u^3 −v^3 =−q and x=u−v (i) will become u^3 +q−((p/(3u)))^3 =0 or (u^3 )^2 +q(u^3 )−((p/3))^3 =0 ..(ii) ⇒u^3 =((−q±(√(q^2 +4((p/3))^3 )))/2) =−(q/2)±(√(((q/2))^2 +((p/3))^3 )) how the solutions will be depends on Δ=((q/2))^2 +((p/3))^3 one case is e.g.: ⇒u=^3 (√(−(q/2)±(√(((q/2))^2 +((p/3))^3 )))) v^3 =u^3 +q=−(q/2)±(√(((q/2))^2 +((p/3))^3 ))+q v^3 =(q/2)±(√(((q/2))^2 +((p/3))^3 )) ⇒v=^3 (√((q/2)±(√(((q/2))^2 +((p/3))^3 )))) x=u−v ⇒x=^3 (√(−(q/2)±(√(((q/2))^2 +((p/3))^3 ))))−^3 (√((q/2)±(√(((q/2))^2 +((p/3))^3 ))))

$${Yes},\:{every}\:{cubic}\:{equation}\:{can}\:{be} \\ $$$${reduced}\:{to}\:{the}\:{form} \\ $$$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\:\:\:...\left({i}\right) \\ $$$${let}\:{v}=\frac{{p}}{\mathrm{3}{u}},\:{u}^{\mathrm{3}} −{v}^{\mathrm{3}} =−{q} \\ $$$${and}\:{x}={u}−{v} \\ $$$$\left({i}\right)\:{will}\:{become} \\ $$$${u}^{\mathrm{3}} +{q}−\left(\frac{{p}}{\mathrm{3}{u}}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$${or} \\ $$$$\left({u}^{\mathrm{3}} \right)^{\mathrm{2}} +{q}\left({u}^{\mathrm{3}} \right)−\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} =\mathrm{0}\:\:..\left({ii}\right) \\ $$$$\Rightarrow{u}^{\mathrm{3}} =\frac{−{q}\pm\sqrt{{q}^{\mathrm{2}} +\mathrm{4}\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} }}{\mathrm{2}} \\ $$$$=−\frac{{q}}{\mathrm{2}}\pm\sqrt{\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$$${how}\:{the}\:{solutions}\:{will}\:{be}\:{depends}\:{on} \\ $$$$\Delta=\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$$ \\ $$$${one}\:{case}\:{is}\:{e}.{g}.: \\ $$$$\Rightarrow{u}=\:^{\mathrm{3}} \sqrt{−\frac{{q}}{\mathrm{2}}\pm\sqrt{\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} }} \\ $$$${v}^{\mathrm{3}} ={u}^{\mathrm{3}} +{q}=−\frac{{q}}{\mathrm{2}}\pm\sqrt{\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} }+{q} \\ $$$${v}^{\mathrm{3}} =\frac{{q}}{\mathrm{2}}\pm\sqrt{\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow{v}=\:^{\mathrm{3}} \sqrt{\frac{{q}}{\mathrm{2}}\pm\sqrt{\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} }} \\ $$$$ \\ $$$${x}={u}−{v} \\ $$$$\Rightarrow{x}=\:^{\mathrm{3}} \sqrt{−\frac{{q}}{\mathrm{2}}\pm\sqrt{\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} }}−\:^{\mathrm{3}} \sqrt{\frac{{q}}{\mathrm{2}}\pm\sqrt{\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} }} \\ $$

Commented by ajfour last updated on 23/Dec/17

great fortune sir. thank you very sincerely!

$${great}\:{fortune}\:{sir}.\:{thank}\:{you} \\ $$$${very}\:{sincerely}! \\ $$

Commented by ajfour last updated on 10/Jun/18

my favourite post.

$${my}\:{favourite}\:{post}. \\ $$

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