Question Number 24565 by ajfour last updated on 21/Nov/17
![y=ax^3 +bx^2 +cx+d , then prove that the equation y=0 has only one real root if a[(9ad−bc)^2 −4(b^2 −3ac)(c^2 −3bd)] > 0 provided b^2 > 3ac .](Q24565.png)
$$\:\:\boldsymbol{{y}}=\boldsymbol{{ax}}^{\mathrm{3}} +\boldsymbol{{bx}}^{\mathrm{2}} +\boldsymbol{{cx}}+\boldsymbol{{d}}\:,\:{then} \\ $$ $${prove}\:{that}\:{the}\:{equation}\:{y}=\mathrm{0} \\ $$ $${has}\:{only}\:{one}\:{real}\:{root}\:{if} \\ $$ $$\:\boldsymbol{{a}}\left[\left(\mathrm{9}\boldsymbol{{ad}}−\boldsymbol{{bc}}\right)^{\mathrm{2}} −\mathrm{4}\left(\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{ac}}\right)\left(\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{bd}}\right)\right] \\ $$ $$\:\:\:\:>\:\mathrm{0}\:\:\:\:\:{provided}\:\:\:\boldsymbol{{b}}^{\mathrm{2}} \:>\:\mathrm{3}\boldsymbol{{ac}}\:. \\ $$
Answered by ajfour last updated on 21/Nov/17
Commented byajfour last updated on 21/Nov/17
![If the local minimum value and the local maximum value, both, are of the same sign, then, i believe, there can be just one real root of a cubic equation. y=ax^3 +bx^2 +cx+d ⇒ (dy/dx)=3ax^2 +2bx+c let at x= α, β (dy/dx)=0 ⇒ 𝛂𝛃=(c/(3a)) and (𝛂+𝛃)=−((2b)/(3a)) ⇒ 3aα^2 +2bα+c =0 .....(i) 3aβ^( 2) +2bβ+c =0 .....(ii) For one real root y(𝛂)×y(𝛃) > 0 or 3y(𝛂)×3y(𝛃) > 0 3y(α)= 3aα^3 +3bα^2 +3cα+3d subtracting α×(i) from this 3y(α)=b𝛂^2 +2c𝛂+3d =(b/(3a))(3a𝛂^2 )+2c𝛂+3d using (i) again: 3y(α)=−(b/(3a))(2bα+c)+2cα+3d =2α(c−(b^2 /(3a)))+(3d−((bc)/(3a))) so 3y(α)×3y(β) = [4𝛂𝛃(c−(b^2 /(3a)))^2 +2(𝛂+𝛃)(c−(b^2 /(3a)))(3d−((bc)/(3a))) +(3d−((bc)/(3a)))^2 ] As α=(c/(3a)) and β=−((2b)/(3a)) we have 3y(α)×3y(β)= ((4c)/(3a))(c−(b^2 /(3a)))^2 −((4b)/(3a))(c−(b^2 /(3a)))(3d−((bc)/(3a))) +(3d−((bc)/(3a)))^2 > 0 or 4c(3ac−b^2 )^2 −4b(3ac−b^2 )(9ad−bc) +3a(9ad−bc)^2 > 0 or 3a(9ad−bc)^2 +4(3ac−b^2 )(3ac^2 −b^2 c −9abd+b^2 c) > 0 ⇒ 3a(9ad−bc)^2 +4(3a)(3ac−b^2 )(c^2 −3bd)>0 a[(9ad−bc)^2 −4(b^2 −3ac)(c^2 −3bd)]>0 .](Q24581.png)
$${If}\:{the}\:{local}\:{minimum}\:{value} \\ $$ $${and}\:{the}\:{local}\:{maximum}\:{value}, \\ $$ $${both},\:{are}\:{of}\:{the}\:{same}\:{sign},\:{then}, \\ $$ $${i}\:{believe},\:{there}\:{can}\:{be}\:{just}\:{one} \\ $$ $${real}\:{root}\:{of}\:{a}\:{cubic}\:{equation}. \\ $$ $$\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{ax}}^{\mathrm{3}} +\boldsymbol{{bx}}^{\mathrm{2}} +\boldsymbol{{cx}}+\boldsymbol{{d}} \\ $$ $$\Rightarrow\:\:\frac{{dy}}{{dx}}=\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c}\: \\ $$ $${let}\:\:{at}\:{x}=\:\alpha,\:\beta\:\:\:\:\:\frac{{dy}}{{dx}}=\mathrm{0} \\ $$ $$\Rightarrow\:\boldsymbol{\alpha\beta}=\frac{\boldsymbol{{c}}}{\mathrm{3}\boldsymbol{{a}}}\:\:{and}\:\:\:\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)=−\frac{\mathrm{2}\boldsymbol{{b}}}{\mathrm{3}\boldsymbol{{a}}} \\ $$ $$\Rightarrow\:\:\:\mathrm{3}{a}\alpha^{\mathrm{2}} +\mathrm{2}{b}\alpha+{c}\:=\mathrm{0}\:\:.....\left({i}\right) \\ $$ $$\:\:\:\:\:\:\:\:\mathrm{3}{a}\beta^{\:\mathrm{2}} +\mathrm{2}{b}\beta+{c}\:=\mathrm{0}\:\:\:\:.....\left({ii}\right) \\ $$ $${For}\:{one}\:{real}\:{root} \\ $$ $$\boldsymbol{{y}}\left(\boldsymbol{\alpha}\right)×\boldsymbol{{y}}\left(\boldsymbol{\beta}\right)\:>\:\mathrm{0} \\ $$ $${or}\:\:\mathrm{3}\boldsymbol{{y}}\left(\boldsymbol{\alpha}\right)×\mathrm{3}\boldsymbol{{y}}\left(\boldsymbol{\beta}\right)\:>\:\mathrm{0} \\ $$ $$\mathrm{3}{y}\left(\alpha\right)=\:\:\mathrm{3}{a}\alpha^{\mathrm{3}} +\mathrm{3}{b}\alpha^{\mathrm{2}} +\mathrm{3}{c}\alpha+\mathrm{3}{d}\: \\ $$ $${subtracting}\:\alpha×\left({i}\right)\:{from}\:{this} \\ $$ $$\mathrm{3}{y}\left(\alpha\right)=\boldsymbol{{b}\alpha}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{c}\alpha}+\mathrm{3}\boldsymbol{{d}} \\ $$ $$\:\:\:\:\:\:\:\:\:=\frac{\boldsymbol{{b}}}{\mathrm{3}\boldsymbol{{a}}}\left(\mathrm{3}\boldsymbol{{a}\alpha}^{\mathrm{2}} \right)+\mathrm{2}\boldsymbol{{c}\alpha}+\mathrm{3}\boldsymbol{{d}} \\ $$ $${using}\:\left({i}\right)\:{again}: \\ $$ $$\:\:\mathrm{3}{y}\left(\alpha\right)=−\frac{{b}}{\mathrm{3}{a}}\left(\mathrm{2}{b}\alpha+{c}\right)+\mathrm{2}{c}\alpha+\mathrm{3}{d} \\ $$ $$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\alpha\left({c}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}\right)+\left(\mathrm{3}{d}−\frac{{bc}}{\mathrm{3}{a}}\right) \\ $$ $${so}\:\:\:\mathrm{3}{y}\left(\alpha\right)×\mathrm{3}{y}\left(\beta\right)\:= \\ $$ $$\:\:\:\:\left[\mathrm{4}\boldsymbol{\alpha\beta}\left(\boldsymbol{{c}}−\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\mathrm{3}\boldsymbol{{a}}}\right)^{\mathrm{2}} +\mathrm{2}\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)\left(\boldsymbol{{c}}−\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\mathrm{3}\boldsymbol{{a}}}\right)\left(\mathrm{3}\boldsymbol{{d}}−\frac{\boldsymbol{{bc}}}{\mathrm{3}\boldsymbol{{a}}}\right)\right. \\ $$ $$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{3}\boldsymbol{{d}}−\frac{\boldsymbol{{bc}}}{\mathrm{3}\boldsymbol{{a}}}\right)^{\mathrm{2}} \right] \\ $$ $$\:{As}\:\:\alpha=\frac{{c}}{\mathrm{3}{a}}\:\:\:{and}\:\:\beta=−\frac{\mathrm{2}{b}}{\mathrm{3}{a}}\:\:\:{we}\:{have} \\ $$ $$\mathrm{3}{y}\left(\alpha\right)×\mathrm{3}{y}\left(\beta\right)= \\ $$ $$\:\:\:\frac{\mathrm{4}{c}}{\mathrm{3}{a}}\left({c}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}\right)^{\mathrm{2}} −\frac{\mathrm{4}{b}}{\mathrm{3}{a}}\left({c}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}\right)\left(\mathrm{3}{d}−\frac{{bc}}{\mathrm{3}{a}}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{3}{d}−\frac{{bc}}{\mathrm{3}{a}}\right)^{\mathrm{2}} \:>\:\mathrm{0} \\ $$ $${or} \\ $$ $$\mathrm{4}{c}\left(\mathrm{3}{ac}−{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{b}\left(\mathrm{3}{ac}−{b}^{\mathrm{2}} \right)\left(\mathrm{9}{ad}−{bc}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}{a}\left(\mathrm{9}{ad}−{bc}\right)^{\mathrm{2}} \:>\:\mathrm{0} \\ $$ $${or} \\ $$ $$\mathrm{3}{a}\left(\mathrm{9}{ad}−{bc}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{3}{ac}−{b}^{\mathrm{2}} \right)\left(\mathrm{3}{ac}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}\right. \\ $$ $$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{9}{abd}+{b}^{\mathrm{2}} {c}\right)\:>\:\mathrm{0} \\ $$ $$\Rightarrow \\ $$ $$\mathrm{3}{a}\left(\mathrm{9}{ad}−{bc}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{3}{a}\right)\left(\mathrm{3}{ac}−{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −\mathrm{3}{bd}\right)>\mathrm{0} \\ $$ $${a}\left[\left(\mathrm{9}{ad}−{bc}\right)^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} −\mathrm{3}{ac}\right)\left({c}^{\mathrm{2}} −\mathrm{3}{bd}\right)\right]>\mathrm{0}\:. \\ $$