Question Number 197753 by tri26112004 last updated on 27/Sep/23 | ||
$${Solve}\:{the}\:{equation}: \\ $$$$\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}}−\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{20}}=\mathrm{5}\sqrt{{x}+\mathrm{1}} \\ $$ | ||
Answered by Frix last updated on 27/Sep/23 | ||
$$\mathrm{Squaring},\:\mathrm{transforming},\:\mathrm{solving},\:\mathrm{testing} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} =\mathrm{8} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{5}+\sqrt{\mathrm{61}}}{\mathrm{2}} \\ $$ | ||
Commented by tri26112004 last updated on 27/Sep/23 | ||
$${can}\:{you}\:{show}\:{your}\:{solution} \\ $$ | ||
Commented by Frix last updated on 27/Sep/23 | ||
$$\sqrt{{a}}−\sqrt{{b}}=\sqrt{{c}} \\ $$$$\sqrt{{a}}=\sqrt{{b}}+\sqrt{{c}} \\ $$$${a}={b}+\mathrm{2}\sqrt{{bc}}+{c} \\ $$$${a}−\left({b}+{c}\right)=\mathrm{2}\sqrt{{bc}} \\ $$$$\left({a}−\left({b}+{c}\right)\right)^{\mathrm{2}} =\mathrm{4}{bc} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}\left({ab}+{ac}+{bc}\right)=\mathrm{0} \\ $$$$\mathrm{Now}\:\mathrm{insert} \\ $$$${a}=\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9} \\ $$$${b}={x}^{\mathrm{2}} −{x}−\mathrm{20} \\ $$$${c}=\mathrm{25}\left({x}+\mathrm{1}\right) \\ $$$${x}^{\mathrm{4}} −\frac{\mathrm{45}{x}^{\mathrm{3}} }{\mathrm{4}}+\frac{\mathrm{33}{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{505}{x}}{\mathrm{4}}+\mathrm{126}=\mathrm{0} \\ $$$$\left({x}−\mathrm{8}\right)\left({x}+\frac{\mathrm{7}}{\mathrm{4}}\right)\left({x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{9}\right)=\mathrm{0} \\ $$$$\mathrm{Test}\:\mathrm{the}\:\mathrm{solutions},\:\mathrm{some}\:\mathrm{might}\:\mathrm{be}\:\mathrm{false}\:\mathrm{due} \\ $$$$\mathrm{to}\:\mathrm{squaring} \\ $$ | ||
Commented by tri26112004 last updated on 28/Sep/23 | ||
$${Do}\:{you}\:{have}\:{another}\:{solution}¿ \\ $$ | ||
Commented by tri26112004 last updated on 28/Sep/23 | ||
$${Such}\:{as}\:{Using}\:{Inequality} \\ $$ | ||
Answered by Rasheed.Sindhi last updated on 28/Sep/23 | ||
$$\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}}−\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{20}}=\mathrm{5}\sqrt{{x}+\mathrm{1}} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{20}\geqslant\mathrm{0}\:\wedge\:\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}\geqslant\mathrm{0}\:\wedge\:{x}+\mathrm{1}\geqslant\mathrm{0} \\ $$$$\left({x}−\mathrm{5}\right)\left({x}+\mathrm{4}\right)\geqslant\mathrm{0} \\ $$$${x}−\mathrm{5}\geqslant\:\mathrm{0}\:\wedge\:{x}+\mathrm{4}\geqslant\mathrm{0} \\ $$$${x}\geqslant\mathrm{5}\:\wedge\:{x}\geqslant−\mathrm{4}\Rightarrow{x}\geqslant\mathrm{5}.......\left({i}\right) \\ $$$$ \\ $$$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}\geqslant\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left(\mathrm{5}{x}+\mathrm{9}\right)\geqslant\mathrm{0} \\ $$$${x}\geqslant−\mathrm{1}\wedge{x}\geqslant−\frac{\mathrm{9}}{\mathrm{5}}\Rightarrow{x}\geqslant−\mathrm{1}....\left({ii}\right) \\ $$$${x}+\mathrm{1}\geqslant\mathrm{0}\Rightarrow{x}\geqslant−\mathrm{1}............\left({iii}\right) \\ $$$$ \\ $$$$\left({i}\right),\left({ii}\right)\:\&\:\left({iii}\right): \\ $$$${x}\geqslant\mathrm{5}\:\wedge\:{x}\geqslant−\mathrm{1} \\ $$$$\Rightarrow\:\:\:{x}\geqslant\mathrm{5} \\ $$$$ \\ $$$$\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}}\:>\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{20}} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}\:>{x}^{\mathrm{2}} −{x}−\mathrm{20} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{29}>\mathrm{0} \\ $$$$ \\ $$ | ||
Commented by tri26112004 last updated on 28/Sep/23 | ||
$${Can}\:{you}\:{use}\:{the}\:{inequality}¿ \\ $$ | ||
Commented by Rasheed.Sindhi last updated on 28/Sep/23 | ||
$${I}'{ve}\:{tried}\:{that}\:{but}\:{failed}\:{to}\:{get}\:{exact}\:{answer}. \\ $$ | ||
Commented by tri26112004 last updated on 28/Sep/23 | ||
$${But}\:{it}\:{has}\:{some}\:{beauty}\:{root} \\ $$$${x}=\mathrm{8}... \\ $$ | ||
Commented by Frix last updated on 28/Sep/23 | ||
$$\mathrm{The}\:\mathrm{solution}\:\mathrm{method}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{care}\:\mathrm{whether} \\ $$$$\mathrm{you}\:\mathrm{like}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{or}\:\mathrm{not}. \\ $$ | ||
Commented by tri26112004 last updated on 28/Sep/23 | ||
$${okay} \\ $$ | ||
Answered by ajfour last updated on 28/Sep/23 | ||
$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}=\mathrm{5}\left({x}^{\mathrm{2}} −{x}−\mathrm{20}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{19}\left({x}+\mathrm{1}\right)+\mathrm{90} \\ $$$$\sqrt{{a}}−\sqrt{{b}}=\mathrm{5}\sqrt{{c}} \\ $$$${a}=\mathrm{5}{b}+\mathrm{19}{c}+\mathrm{90} \\ $$$${b}={c}^{\mathrm{2}} −\mathrm{3}{c}−\mathrm{18} \\ $$$$\Rightarrow\:\:{a}−{b}=\mathrm{4}{b}+\mathrm{19}{c}+\mathrm{90} \\ $$$${Also}\:\:\:{a}=\mathrm{5}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\left({x}+\mathrm{1}\right) \\ $$$$\sqrt{{a}}+\sqrt{{b}}=\frac{\mathrm{4}{b}+\mathrm{19}{c}+\mathrm{90}}{\mathrm{5}\sqrt{{c}}} \\ $$$$\mathrm{2}\sqrt{{a}}=\:\frac{\mathrm{25}{c}+\mathrm{4}{b}+\mathrm{19}{c}+\mathrm{90}}{\mathrm{5}\sqrt{{c}}} \\ $$$$\mathrm{100}{ac}=\left(\mathrm{4}{b}+\mathrm{44}{c}+\mathrm{90}\right)^{\mathrm{2}} \\ $$$$\mathrm{100}\left(\mathrm{5}{c}^{\mathrm{2}} +\mathrm{4}{c}\right){c}=\left\{\mathrm{4}\left({c}^{\mathrm{2}} −\mathrm{3}{c}−\mathrm{18}\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{44}{c}+\mathrm{90}\right\}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{500}{c}^{\mathrm{3}} +\mathrm{400}{c}^{\mathrm{2}} \\ $$$$\:\:\:\:=\left(\mathrm{4}{c}^{\mathrm{2}} +\mathrm{32}{c}+\mathrm{18}\right)^{\mathrm{2}} \\ $$$$\mathrm{25}{c}^{\mathrm{2}} \left(\mathrm{5}{c}+\mathrm{4}\right)=\left(\mathrm{2}{c}^{\mathrm{2}} +\mathrm{16}{c}+\mathrm{9}\right)^{\mathrm{2}} \\ $$$${this}\:{need}\:{to}\:{be}\:{solved} \\ $$$${Thereafter}\:\:\:\:{as}\:\:\:{x}+\mathrm{1}={c} \\ $$$${x}={c}−\mathrm{1} \\ $$$$ \\ $$ | ||