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