Question Number 79279 by jagoll last updated on 24/Jan/20
$$\mathrm{solve}\: \\ $$$$\mid\mathrm{x}\mid^{\mathrm{3}} −\mathrm{7x}^{\mathrm{2}} +\mathrm{7}\mid\mathrm{x}\mid+\mathrm{15}<\mathrm{0} \\ $$
Commented by jagoll last updated on 24/Jan/20
$$\mathrm{thanks}\: \\ $$
Commented by MJS last updated on 24/Jan/20
$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\ $$
Commented by john santu last updated on 24/Jan/20
$${where}\:{is}\:{wrong}? \\ $$
Commented by MJS last updated on 24/Jan/20
$$\mathrm{you}\:\mathrm{should}\:\mathrm{plot}\:\mathrm{the}\:\mathrm{function}\:\mathrm{or}\:\mathrm{scetch}\:\mathrm{it} \\ $$
Answered by MJS last updated on 24/Jan/20
$$\mathrm{case}\:\mathrm{1}\:{x}\geqslant\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{7}{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{15}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=−\mathrm{1}\vee{x}=\mathrm{3}\vee{x}=\mathrm{5} \\ $$$$\mathrm{testing}\:{x}=\mathrm{4}:\:\mathrm{4}^{\mathrm{3}} −\mathrm{7}×\mathrm{4}^{\mathrm{2}} +\mathrm{7}×\mathrm{4}+\mathrm{15}=−\mathrm{5} \\ $$$$\mathrm{case}\:\mathrm{2}\:{x}<\mathrm{0} \\ $$$$−{x}^{\mathrm{3}} −\mathrm{7}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{15}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=−\mathrm{5}\vee{x}=−\mathrm{3}\vee{x}=\mathrm{1} \\ $$$$\mathrm{testing}\:{x}=−\mathrm{4}:\:−\left(−\mathrm{4}\right)^{\mathrm{3}} −\mathrm{7}×\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{7}×\mathrm{4}+\mathrm{15}=−\mathrm{5} \\ $$$$ \\ $$$$\Rightarrow \\ $$$$−\mathrm{5}<{x}<−\mathrm{3}\vee\mathrm{3}<{x}<\mathrm{5} \\ $$$$\mathrm{or} \\ $$$$\mathrm{3}<\mid{x}\mid<\mathrm{5} \\ $$
Commented by john santu last updated on 24/Jan/20
$${your}\:{are}\:{right}.\:{i}'{m}\:{forgot}\: \\ $$$${the}\:{definition}\:{sign}\left({x}\right) \\ $$
Commented by jagoll last updated on 24/Jan/20
$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$