Question Number 50278 by LYCON TRIX last updated on 15/Dec/18
$$\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}\:+\:\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}\:=\:\mathrm{2x} \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{beg}\:\mathrm{u}\:\mathrm{guys}\: \\ $$$$\mathrm{please}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{question} \\ $$
Commented by ajfour last updated on 15/Dec/18
$${a}\:{and}\:{x}\:\in\:\mathbb{R}\:\:{or}\:\mathbb{C}\:? \\ $$
Commented by MJS last updated on 15/Dec/18
$$\mathrm{would}\:\mathrm{you}\:\mathrm{be}\:\mathrm{so}\:\mathrm{kind}\:\mathrm{to}\:\mathrm{tell}\:\mathrm{us}\:\mathrm{where}\:\mathrm{this} \\ $$$$\mathrm{equation}\:\mathrm{comes}\:\mathrm{from}? \\ $$
Commented by maxmathsup by imad last updated on 15/Dec/18
$${its}\:{only}\:{a}\:{trying}\:{let}\:{use}\:{the}\:{changement}\:{x}\:={acos}\left(\mathrm{2}{t}\right) \\ $$$$\left({e}\right)\:\Leftrightarrow\:\sqrt{{a}+\sqrt{{a}−{acos}\left(\mathrm{2}{t}\right)}}+\sqrt{{a}−\sqrt{{a}}}=\mathrm{2}{a}\:{cos}\left(\mathrm{2}{t}\right)\:\Rightarrow \\ $$$$\sqrt{{a}+\sqrt{{a}}\sqrt{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}}+\sqrt{{a}−\sqrt{{a}}\sqrt{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}}=\mathrm{2}{a}\:{cos}\left(\mathrm{2}{t}\right)\:\Rightarrow \\ $$$$\sqrt{{a}+\sqrt{\mathrm{2}{a}}{sint}}+\sqrt{{a}−\sqrt{\mathrm{2}{a}}{cost}}=\mathrm{2}{acos}\left(\mathrm{2}{t}\right)\:\:\Rightarrow \\ $$$${a}+\sqrt{\mathrm{2}{a}}{sint}\:+\mathrm{2}\sqrt{{a}+\sqrt{\mathrm{2}{a}}{sint}}\sqrt{{a}−\sqrt{\mathrm{2}{a}}{cost}}\:+{a}−\sqrt{\mathrm{2}{a}}{cost}\:=\mathrm{4}{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\:\Rightarrow \\ $$$$\left.\mathrm{2}{a}+\sqrt{\mathrm{2}{a}}\left({sint}\:−{cost}\right)\:+\mathrm{2}\sqrt{\left({a}+\sqrt{\mathrm{2}{a}}{sint}\right)\left({a}−\sqrt{\mathrm{2}{a}}{cost}\right.}\right)\:=\mathrm{4}{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\:\Rightarrow \\ $$$$\left(\mathrm{4}{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)−\mathrm{2}{a}−\sqrt{\mathrm{2}{a}}\left({sint}\:−{cost}\right)\right)^{\mathrm{2}} =\mathrm{4}\left({a}^{\mathrm{2}} −{a}\sqrt{\mathrm{2}{a}}{cost}\:+{a}\sqrt{\mathrm{2}{a}}{sint}\:−\mathrm{2}{asint}\:{cost}\right) \\ $$$$….{be}\:{cntinued}\:\:\:{i}\:{think}\:{that}\:{way}\:{can}\:{give}\:{something}\:{if}\:{we}\:{resist}\:{to}\:{storm} \\ $$$${of}\:{calculus}… \\ $$
Commented by ajfour last updated on 23/Aug/23
$${A}\:{past}\:{prize}\:{question},\:{i}\:{wonder}\:{who}\:{got}\:{the}\:{prize}. \\ $$$${just}\:{name}\:{the}\:{guy},\:{wont}\:{u}? \\ $$
Answered by behi83417@gmail.com last updated on 17/Dec/18
$${a}+{x}={t}^{\mathrm{2}} ,{a}−{x}={s}^{\mathrm{2}} \Rightarrow{x}=\frac{{t}^{\mathrm{2}} −{s}^{\mathrm{2}} }{\mathrm{2}},{a}=\frac{{t}^{\mathrm{2}} +{s}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\sqrt{{a}+{s}}+\sqrt{{a}−{t}}={t}^{\mathrm{2}} −{s}^{\mathrm{2}} \\ $$$${a}+{s}+{a}−{t}+\mathrm{2}\sqrt{\left({a}+{s}\right)\left({a}−{t}\right)}=\left({t}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{\left({a}+{s}\right)\left({a}−{t}\right)}=\left({t}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)^{\mathrm{2}} +{t}−{s}−\mathrm{2}{a}= \\ $$$$={t}^{\mathrm{4}} +{s}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} −{t}^{\mathrm{2}} −{s}^{\mathrm{2}} +{t}−{s} \\ $$$$\Rightarrow\mathrm{4}\left({a}^{\mathrm{2}} −{at}+{as}−{st}\right)= \\ $$$$={t}^{\mathrm{8}} +{s}^{\mathrm{8}} +\mathrm{4}{t}^{\mathrm{4}} {s}^{\mathrm{4}} +{t}^{\mathrm{4}} +{s}^{\mathrm{4}} +{t}^{\mathrm{2}} +{s}^{\mathrm{2}} + \\ $$$$+\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{6}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{5}} −\mathrm{2}{t}^{\mathrm{4}} {s} \\ $$$$−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{6}} −\mathrm{2}{s}^{\mathrm{4}} {t}^{\mathrm{2}} −\mathrm{2}{s}^{\mathrm{6}} +\mathrm{2}{ts}^{\mathrm{4}} −\mathrm{2}{s}^{\mathrm{5}} +\mathrm{4}{t}^{\mathrm{4}} {s}^{\mathrm{2}} \\ $$$$+\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{3}} {s}^{\mathrm{2}} +\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s} \\ $$$$−\mathrm{2}{s}^{\mathrm{2}} {t}+\mathrm{2}{s}^{\mathrm{3}} −\mathrm{2}{ts} \\ $$$${LHS}=\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)\left({t}−{s}\right)−\mathrm{4}{st}= \\ $$$$={t}^{\mathrm{4}} +{s}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s}−\mathrm{2}{s}^{\mathrm{2}} {t}+\mathrm{2}{s}^{\mathrm{3}} −\mathrm{4}{st} \\ $$$${RHS}={t}^{\mathrm{8}} +{s}^{\mathrm{8}} +\mathrm{6}{t}^{\mathrm{4}} {s}^{\mathrm{4}} +{t}^{\mathrm{4}} +{s}^{\mathrm{4}} +{t}^{\mathrm{2}} +{s}^{\mathrm{2}} \\ $$$$−\mathrm{4}{t}^{\mathrm{6}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{6}} −\mathrm{2}{s}^{\mathrm{6}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{5}} −\mathrm{2}{t}^{\mathrm{4}} {s} \\ $$$$−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{6}} −\mathrm{2}{s}^{\mathrm{4}} {t}^{\mathrm{2}} +\mathrm{2}{ts}^{\mathrm{4}} −\mathrm{2}{s}^{\mathrm{5}} +\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{3}} {s}^{\mathrm{2}} \\ $$$$+\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s}−\mathrm{2}{s}^{\mathrm{2}} {t}+\mathrm{2}{s}^{\mathrm{3}} −\mathrm{2}{ts} \\ $$$$\Rightarrow{t}^{\mathrm{8}} +{s}^{\mathrm{8}} −\mathrm{2}{t}^{\mathrm{6}} −\mathrm{2}{s}^{\mathrm{6}} +\mathrm{2}{t}^{\mathrm{5}} +\mathrm{2}{s}^{\mathrm{5}} \\ $$$$+{t}^{\mathrm{2}} +{s}^{\mathrm{2}} +\mathrm{2}{ts}−\mathrm{4}{t}^{\mathrm{6}} {s}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{4}} {s}−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{6}} \\ $$$$+\mathrm{2}{s}^{\mathrm{4}} {t}^{\mathrm{2}} +\mathrm{2}{ts}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{3}} {s}^{\mathrm{2}} +\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s} \\ $$$$−\mathrm{2}{s}^{\mathrm{2}} {t}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{8}} +{s}^{\mathrm{8}} \right)−\mathrm{2}\left({t}^{\mathrm{6}} +{s}^{\mathrm{6}} \right)+\mathrm{2}\left({t}^{\mathrm{5}} −{s}^{\mathrm{5}} \right)− \\ $$$$−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{2}} \left({t}^{\mathrm{4}} +{s}^{\mathrm{4}} \right)−\mathrm{2}{ts}\left({t}^{\mathrm{3}} −{s}^{\mathrm{3}} \right)+ \\ $$$$+\left(\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{2}} \left({t}−{s}\right)+ \\ $$$$+\mathrm{6}{t}^{\mathrm{4}} {s}^{\mathrm{4}} +\mathrm{2}{ts}=\mathrm{0} \\ $$$${t}−{s}={p},{ts}={q} \\ $$$${t}^{\mathrm{8}} +{s}^{\mathrm{8}} =\left({t}^{\mathrm{4}} −{s}^{\mathrm{4}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} = \\ $$$$=\left({t}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} = \\ $$$$=\left({t}−{s}\right)^{\mathrm{2}} \left({t}+{s}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} = \\ $$$$=\left({t}−{s}\right)^{\mathrm{2}} \left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{4}{ts}\right)\left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{2}{ts}\right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} = \\ $$$$=\left({p}\right)^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{4}{q}\right)\left({p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}+\mathrm{4}{q}^{\mathrm{2}} \right)+\mathrm{2}{q}^{\mathrm{4}} = \\ $$$$=\left({p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}\right)\left({p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}+\mathrm{4}{q}^{\mathrm{2}} \right)+\mathrm{2}{q}^{\mathrm{4}} = \\ $$$$={p}^{\mathrm{8}} +\mathrm{8}{p}^{\mathrm{6}} {q}+\mathrm{20}{p}^{\mathrm{4}} {q}^{\mathrm{2}} +\mathrm{16}{p}^{\mathrm{2}} {q}^{\mathrm{3}} +\mathrm{2}{q}^{\mathrm{4}} \\ $$$${t}^{\mathrm{6}} +{s}^{\mathrm{6}} =\left({t}^{\mathrm{3}} −{s}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{3}} {s}^{\mathrm{3}} =\left({t}−{s}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +{ts}+{s}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{3}} {s}^{\mathrm{3}} = \\ $$$$=\left({p}\right)^{\mathrm{2}} \left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{3}{ts}\right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{3}} {s}^{\mathrm{3}} = \\ $$$$={p}^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{3}{q}\right)^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{3}} = \\ $$$$={p}^{\mathrm{6}} +\mathrm{6}{p}^{\mathrm{4}} {q}+\mathrm{9}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{3}} \\ $$$${t}^{\mathrm{5}} −{s}^{\mathrm{5}} =\left({t}−{s}\right)\left({t}^{\mathrm{4}} +{t}^{\mathrm{3}} {s}+{t}^{\mathrm{2}} {s}^{\mathrm{2}} +{ts}^{\mathrm{3}} +{s}^{\mathrm{4}} \right)= \\ $$$$=\left({t}−{s}\right)\left({t}^{\mathrm{4}} +{s}^{\mathrm{4}} +{ts}\left({t}^{\mathrm{2}} +{ts}+{s}^{\mathrm{2}} \right)\right)= \\ $$$$=\left({t}−{s}\right)\left(\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} +{ts}\left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{3}{ts}\right)=\right. \\ $$$$=\left({t}−{s}\right)\left[\left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{2}{ts}\right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} +{ts}\left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{3}{ts}\right)\right)= \\ $$$$={p}\left[\left(\left({p}^{\mathrm{2}} +\mathrm{2}{q}\right)^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} +{q}\left({p}^{\mathrm{2}} +\mathrm{3}{q}\right)\right)=\right. \\ $$$$={p}\left[\left({p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}+\mathrm{4}{q}^{\mathrm{2}} \right)−\mathrm{2}{q}^{\mathrm{2}} +{qp}^{\mathrm{2}} +\mathrm{3}{pq}\right]= \\ $$$$={p}^{\mathrm{5}} +\mathrm{5}{p}^{\mathrm{2}} {q}+\mathrm{5}{pq}^{\mathrm{2}} \\ $$$${t}^{\mathrm{4}} +{s}^{\mathrm{4}} =\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} =\left[\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{2}{ts}\right]^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} = \\ $$$$=\left[{p}^{\mathrm{2}} +\mathrm{2}{q}\right]^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} ={p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}+\mathrm{2}{q}^{\mathrm{2}} \\ $$$${t}^{\mathrm{3}} −{s}^{\mathrm{3}} =\left({t}−{s}\right)\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} +{ts}\right)={p}^{\mathrm{3}} +\mathrm{3}{pq} \\ $$$${t}^{\mathrm{2}} +{s}^{\mathrm{2}} =\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{2}{ts}={p}^{\mathrm{2}} +\mathrm{2}{q} \\ $$$${after}\:{replacing}\:{and}\:{simplifing}: \\ $$$$\left({p}^{\mathrm{8}} −\mathrm{2}{p}^{\mathrm{6}} +\mathrm{2}{p}^{\mathrm{5}} +{p}^{\mathrm{2}} \right)+\mathrm{16}{p}^{\mathrm{2}} \left({p}^{\mathrm{2}} −\mathrm{1}\right){q}^{\mathrm{2}} \\ $$$$+\left(\mathrm{8}{p}^{\mathrm{6}} −\mathrm{12}{p}^{\mathrm{4}} +\mathrm{8}{p}^{\mathrm{3}} +\mathrm{4}\right){q}=\mathrm{0} \\ $$
Commented by behi83417@gmail.com last updated on 17/Dec/18
$$\Rightarrow{Q}#\mathrm{50511} \\ $$