Question Number 150189 by mathdanisur last updated on 10/Aug/21 | ||
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\left(\mathrm{x}\:-\:\mathrm{3}\right)\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{x}\:−\:\mathrm{2}}\:=\:\mathrm{0} \\ $$ | ||
Commented by Skabetix last updated on 10/Aug/21 | ||
$${x}=\mathrm{3}\:{or}\:{x}=\mathrm{2}\:{or}\:{x}=−\mathrm{1} \\ $$ | ||
Answered by Rasheed.Sindhi last updated on 10/Aug/21 | ||
$$\left(\mathrm{x}\:-\:\mathrm{3}\right)\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{x}\:−\:\mathrm{2}}\:=\:\mathrm{0} \\ $$$$\mathrm{x}−\mathrm{3}=\mathrm{0}\:\vee\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{x}\:−\:\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{3}\:\vee\:\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{3}\:\vee\:\mathrm{x}=\mathrm{2}\:\vee\mathrm{x}=−\mathrm{1} \\ $$ | ||
Commented by mathdanisur last updated on 10/Aug/21 | ||
$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$ | ||