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x-3-2-x-4-x-5-6-x-2-2-Find-x-




Question Number 207206 by hardmath last updated on 09/May/24
(√(x−3 + 2 (√(x−4)))) − (√(x + 5−6 (√(x−2)))) = 2  Find:  x = ?
$$\sqrt{\mathrm{x}−\mathrm{3}\:+\:\mathrm{2}\:\sqrt{\mathrm{x}−\mathrm{4}}}\:−\:\sqrt{\mathrm{x}\:+\:\mathrm{5}−\mathrm{6}\:\sqrt{\mathrm{x}−\mathrm{2}}}\:=\:\mathrm{2} \\ $$$$\mathrm{Find}:\:\:\mathrm{x}\:=\:? \\ $$
Answered by A5T last updated on 09/May/24
((√(x−4))+1)^2 =x−3+2(√(x−4))  ⇒(√(x−3+2(√(x−4))))=(√(x−4))+1  Question⇒(√(x−4))−(√(x+5−6(√(x−2))))=1  ⇒(√(x−4))=−4+3(√(x−2))⇒3(√(x−2))−(√(x−4))=4  ⇒u=(√(x−2));v=(√(x−4))  ⇒u^2 −v^2 =2; 3u−v=4⇒v=3u−4  ⇒u^2 −(3u−4)^2 =2⇒u^2 −9u^2 −16+24u=2  ⇒8u^2 −24u+18=0⇒4u^2 −12u+9=0  ⇒u=(3/2)⇒u^2 +2=x=(9/4)+2=((17)/4)  Substituting x=((17)/4) gives 1≠2,contradiction→←  Hence,no solution exists.
$$\left(\sqrt{{x}−\mathrm{4}}+\mathrm{1}\right)^{\mathrm{2}} ={x}−\mathrm{3}+\mathrm{2}\sqrt{{x}−\mathrm{4}} \\ $$$$\Rightarrow\sqrt{{x}−\mathrm{3}+\mathrm{2}\sqrt{{x}−\mathrm{4}}}=\sqrt{{x}−\mathrm{4}}+\mathrm{1} \\ $$$${Question}\Rightarrow\sqrt{{x}−\mathrm{4}}−\sqrt{{x}+\mathrm{5}−\mathrm{6}\sqrt{{x}−\mathrm{2}}}=\mathrm{1} \\ $$$$\Rightarrow\sqrt{{x}−\mathrm{4}}=−\mathrm{4}+\mathrm{3}\sqrt{{x}−\mathrm{2}}\Rightarrow\mathrm{3}\sqrt{{x}−\mathrm{2}}−\sqrt{{x}−\mathrm{4}}=\mathrm{4} \\ $$$$\Rightarrow{u}=\sqrt{{x}−\mathrm{2}};{v}=\sqrt{{x}−\mathrm{4}} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −{v}^{\mathrm{2}} =\mathrm{2};\:\mathrm{3}{u}−{v}=\mathrm{4}\Rightarrow{v}=\mathrm{3}{u}−\mathrm{4} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −\left(\mathrm{3}{u}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{2}\Rightarrow{u}^{\mathrm{2}} −\mathrm{9}{u}^{\mathrm{2}} −\mathrm{16}+\mathrm{24}{u}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{8}{u}^{\mathrm{2}} −\mathrm{24}{u}+\mathrm{18}=\mathrm{0}\Rightarrow\mathrm{4}{u}^{\mathrm{2}} −\mathrm{12}{u}+\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow{u}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{u}^{\mathrm{2}} +\mathrm{2}={x}=\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{2}=\frac{\mathrm{17}}{\mathrm{4}} \\ $$$${Substituting}\:{x}=\frac{\mathrm{17}}{\mathrm{4}}\:{gives}\:\mathrm{1}\neq\mathrm{2},{contradiction}\rightarrow\leftarrow \\ $$$${Hence},{no}\:{solution}\:{exists}. \\ $$
Commented by hardmath last updated on 09/May/24
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Commented by hardmath last updated on 10/May/24
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Frix last updated on 09/May/24
1. 0≤x−4  ⇒ 4≤x ⇒ 1≤(√(x−3+2(√(x−4))))  2. 0≤x+5−6(√(x−2))  ⇒ 2≤x≤13−6(√2)∨x≥13+6(√2)  =================  ⇒ 4≤x≤13−6(√2)∨x≥13+6(√2)  A. 4≤x≤13−6(√2)  ⇒ 1+(√3)−(√6)≤(√(x−3+2(√(x−4))))−(√(x+5−6(√(x−2))))≤1−(√3)+(√6)       [≈ .283≤lhs≤1.717]  B. 13+6(√2)≤x  ⇒ 4<(√(x−3+2(√(x−4))))−(√(x+5−6(√(x−2))))≤1+(√3)+(√6)       [≈ 4<lhs≤5.182]    ⇒ no solution
$$\mathrm{1}.\:\mathrm{0}\leqslant{x}−\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{4}\leqslant{x}\:\Rightarrow\:\mathrm{1}\leqslant\sqrt{{x}−\mathrm{3}+\mathrm{2}\sqrt{{x}−\mathrm{4}}} \\ $$$$\mathrm{2}.\:\mathrm{0}\leqslant{x}+\mathrm{5}−\mathrm{6}\sqrt{{x}−\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}\leqslant{x}\leqslant\mathrm{13}−\mathrm{6}\sqrt{\mathrm{2}}\vee{x}\geqslant\mathrm{13}+\mathrm{6}\sqrt{\mathrm{2}} \\ $$$$================= \\ $$$$\Rightarrow\:\mathrm{4}\leqslant{x}\leqslant\mathrm{13}−\mathrm{6}\sqrt{\mathrm{2}}\vee{x}\geqslant\mathrm{13}+\mathrm{6}\sqrt{\mathrm{2}} \\ $$$${A}.\:\mathrm{4}\leqslant{x}\leqslant\mathrm{13}−\mathrm{6}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{1}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{6}}\leqslant\sqrt{{x}−\mathrm{3}+\mathrm{2}\sqrt{{x}−\mathrm{4}}}−\sqrt{{x}+\mathrm{5}−\mathrm{6}\sqrt{{x}−\mathrm{2}}}\leqslant\mathrm{1}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}} \\ $$$$\:\:\:\:\:\left[\approx\:.\mathrm{283}\leqslant\mathrm{lhs}\leqslant\mathrm{1}.\mathrm{717}\right] \\ $$$${B}.\:\mathrm{13}+\mathrm{6}\sqrt{\mathrm{2}}\leqslant{x} \\ $$$$\Rightarrow\:\mathrm{4}<\sqrt{{x}−\mathrm{3}+\mathrm{2}\sqrt{{x}−\mathrm{4}}}−\sqrt{{x}+\mathrm{5}−\mathrm{6}\sqrt{{x}−\mathrm{2}}}\leqslant\mathrm{1}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}} \\ $$$$\:\:\:\:\:\left[\approx\:\mathrm{4}<\mathrm{lhs}\leqslant\mathrm{5}.\mathrm{182}\right] \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$
Commented by hardmath last updated on 10/May/24
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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