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Question-210895

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Question Number 210895 by RojaTaniya last updated on 21/Aug/24
Answered by Rasheed.Sindhi last updated on 21/Aug/24
x−(√((10)/x)) =11∧ x∈R⇒x>0 x−(√((10x)/x^2 )) =11 x−((√(10x))/(∣x∣))=11 For x>0 x^2 −11x=(√(10x)) ⇒ determinant ((((√(10x)) =x^2 −11x))) {x(x−11)}^2 =10x x^2 (x^2 −22x+121)=10x x^4 −22x^3 +121x^2 −10x=0 x^3 −22x^2 +121x−10=0 ; x≠0 { ((x=10 Invalid)),((x=6+(√(35)) ✓)),((x=6−(√(35)) Invalid)) :}⇒ determinant (((x=6+(√(35))))) ▶x−(√(10x)) =x−x^2 +11x=−x^2 +12x =x(12−x) =(6+(√(35)) )(12−6−(√(35)) ) =(6+(√(35)) )(6−(√(35)) ) =6^2 −35=1 ✓
$${x}−\sqrt{\frac{\mathrm{10}}{{x}}}\:=\mathrm{11}\wedge\:{x}\in\mathbb{R}\Rightarrow{x}>\mathrm{0} \\ $$$$\: \\ $$$${x}−\sqrt{\frac{\mathrm{10}{x}}{{x}^{\mathrm{2}} }}\:=\mathrm{11} \\ $$$${x}−\frac{\sqrt{\mathrm{10}{x}}}{\mid{x}\mid}=\mathrm{11} \\ $$$${For}\:{x}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{11}{x}=\sqrt{\mathrm{10}{x}}\:\Rightarrow\begin{array}{|c|}{\sqrt{\mathrm{10}{x}}\:={x}^{\mathrm{2}} −\mathrm{11}{x}}\\\hline\end{array} \\ $$$$\:\left\{{x}\left({x}−\mathrm{11}\right)\right\}^{\mathrm{2}} =\mathrm{10}{x} \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{121}\right)=\mathrm{10}{x} \\ $$$${x}^{\mathrm{4}} −\mathrm{22}{x}^{\mathrm{3}} +\mathrm{121}{x}^{\mathrm{2}} −\mathrm{10}{x}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{22}{x}^{\mathrm{2}} +\mathrm{121}{x}−\mathrm{10}=\mathrm{0}\:;\:{x}\neq\mathrm{0} \\ $$$$\begin{cases}{{x}=\mathrm{10}\:{Invalid}}\\{{x}=\mathrm{6}+\sqrt{\mathrm{35}}\:\checkmark}\\{{x}=\mathrm{6}−\sqrt{\mathrm{35}}\:\:{Invalid}}\end{cases}\Rightarrow\begin{array}{|c|}{{x}=\mathrm{6}+\sqrt{\mathrm{35}}}\\\hline\end{array} \\ $$$$\: \\ $$$$\blacktriangleright{x}−\sqrt{\mathrm{10}{x}}\:={x}−{x}^{\mathrm{2}} +\mathrm{11}{x}=−{x}^{\mathrm{2}} +\mathrm{12}{x} \\ $$$$\:\:\:\:\:\:\:\:\:={x}\left(\mathrm{12}−{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\mathrm{6}+\sqrt{\mathrm{35}}\:\right)\left(\mathrm{12}−\mathrm{6}−\sqrt{\mathrm{35}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\mathrm{6}+\sqrt{\mathrm{35}}\:\right)\left(\mathrm{6}−\sqrt{\mathrm{35}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{6}^{\mathrm{2}} −\mathrm{35}=\mathrm{1}\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$
Answered by Frix last updated on 21/Aug/24
x=p>0 (√x)=q>0 { ((p−((√(10))/q)=11 ⇒ q=((√(10))/(p−11)))),((p−(√(10))q=a ⇒ q=(((p−a))/( (√(10)))))) :} ((√(10))/(p−11))=(((p−a))/( (√(10)))) p^2 −(a+11)p+11a−10=0 p=((a+11+s(√(a^2 −22a+161)))/2); s=±1 q=(((p−a)(√(10)))/(10))=−((a−11−s(√(a^2 −22a+161)))/(2(√(10)))) p−q^2 =0 ⇒ s(a−1)s(√(a^2 −22a+161))=(a−31)(a−1) a=1
$${x}={p}>\mathrm{0} \\ $$$$\sqrt{{x}}={q}>\mathrm{0} \\ $$$$\begin{cases}{{p}−\frac{\sqrt{\mathrm{10}}}{{q}}=\mathrm{11}\:\Rightarrow\:{q}=\frac{\sqrt{\mathrm{10}}}{{p}−\mathrm{11}}}\\{{p}−\sqrt{\mathrm{10}}{q}={a}\:\Rightarrow\:{q}=\frac{\left({p}−{a}\right)}{\:\sqrt{\mathrm{10}}}}\end{cases} \\ $$$$\frac{\sqrt{\mathrm{10}}}{{p}−\mathrm{11}}=\frac{\left({p}−{a}\right)}{\:\sqrt{\mathrm{10}}} \\ $$$${p}^{\mathrm{2}} −\left({a}+\mathrm{11}\right){p}+\mathrm{11}{a}−\mathrm{10}=\mathrm{0} \\ $$$${p}=\frac{{a}+\mathrm{11}+{s}\sqrt{{a}^{\mathrm{2}} −\mathrm{22}{a}+\mathrm{161}}}{\mathrm{2}};\:{s}=\pm\mathrm{1} \\ $$$${q}=\frac{\left({p}−{a}\right)\sqrt{\mathrm{10}}}{\mathrm{10}}=−\frac{{a}−\mathrm{11}−{s}\sqrt{{a}^{\mathrm{2}} −\mathrm{22}{a}+\mathrm{161}}}{\mathrm{2}\sqrt{\mathrm{10}}} \\ $$$${p}−{q}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow \\ $$$${s}\left({a}−\mathrm{1}\right){s}\sqrt{{a}^{\mathrm{2}} −\mathrm{22}{a}+\mathrm{161}}=\left({a}−\mathrm{31}\right)\left({a}−\mathrm{1}\right) \\ $$$${a}=\mathrm{1} \\ $$
Answered by Frix last updated on 21/Aug/24
x>0∧(√x)>0 x+((√(10))/( (√x)))=11 ((√x))^3 −11(√x)−(√(10))=0 ((√x)+(√(10)))_(impossible) (((√x))^2 −(√(10))(√x)−1)=0 ((√x)+(((√(14))−(√(10)))/2))_(impossible) ((√x)−(((√(14))+(√(10)))/2))=0 (√x)=(((√(14))+(√(10)))/2) ⇒ x=6+(√(35)) −(√(10x))=−(((√(14))+(√(10)))/2)(√(10))=−5−(√(35)) x−(√(10x))=1
$${x}>\mathrm{0}\wedge\sqrt{{x}}>\mathrm{0} \\ $$$${x}+\frac{\sqrt{\mathrm{10}}}{\:\sqrt{{x}}}=\mathrm{11} \\ $$$$\left(\sqrt{{x}}\right)^{\mathrm{3}} −\mathrm{11}\sqrt{{x}}−\sqrt{\mathrm{10}}=\mathrm{0} \\ $$$$\underset{\mathrm{impossible}} {\underbrace{\left(\sqrt{{x}}+\sqrt{\mathrm{10}}\right)}}\left(\left(\sqrt{{x}}\right)^{\mathrm{2}} −\sqrt{\mathrm{10}}\sqrt{{x}}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\underset{\mathrm{impossible}} {\underbrace{\left(\sqrt{{x}}+\frac{\sqrt{\mathrm{14}}−\sqrt{\mathrm{10}}}{\mathrm{2}}\right)}}\left(\sqrt{{x}}−\frac{\sqrt{\mathrm{14}}+\sqrt{\mathrm{10}}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\sqrt{{x}}=\frac{\sqrt{\mathrm{14}}+\sqrt{\mathrm{10}}}{\mathrm{2}}\:\Rightarrow\:{x}=\mathrm{6}+\sqrt{\mathrm{35}} \\ $$$$−\sqrt{\mathrm{10}{x}}=−\frac{\sqrt{\mathrm{14}}+\sqrt{\mathrm{10}}}{\mathrm{2}}\sqrt{\mathrm{10}}=−\mathrm{5}−\sqrt{\mathrm{35}} \\ $$$${x}−\sqrt{\mathrm{10}{x}}=\mathrm{1} \\ $$
Commented by Frix last updated on 21/Aug/24
x−(√(10x))=((√x)−(√(10)))(√x) x−(√((10)/x))−11=((√x)+(√(10)))((√x)−(1/( (√x)))−(√(10)))=0
$${x}−\sqrt{\mathrm{10}{x}}=\left(\sqrt{{x}}−\sqrt{\mathrm{10}}\right)\sqrt{{x}} \\ $$$${x}−\sqrt{\frac{\mathrm{10}}{{x}}}−\mathrm{11}=\left(\sqrt{{x}}+\sqrt{\mathrm{10}}\right)\left(\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}−\sqrt{\mathrm{10}}\right)=\mathrm{0} \\ $$
Answered by mr W last updated on 21/Aug/24
x−(√((10)/x))=11 let t=(√((10)/x))>0 ⇒x=((10)/t^2 ) ((10)/t^2 )−t=11 ⇒t^3 +11t^2 −10=0 t^3 +t^2 +10t^2 +10t−10t−10=0 (t+1)(t^2 +10t−10)=0 ⇒t^2 +10t−10=0 ⇒10(1−t)=t^2 x−(√(10x))=x(1−(√((10)/x))) =((10(1−t))/t^2 )=1 ✓
$${x}−\sqrt{\frac{\mathrm{10}}{{x}}}=\mathrm{11} \\ $$$${let}\:{t}=\sqrt{\frac{\mathrm{10}}{{x}}}>\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{10}}{{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{10}}{{t}^{\mathrm{2}} }−{t}=\mathrm{11} \\ $$$$\Rightarrow{t}^{\mathrm{3}} +\mathrm{11}{t}^{\mathrm{2}} −\mathrm{10}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} +{t}^{\mathrm{2}} +\mathrm{10}{t}^{\mathrm{2}} +\mathrm{10}{t}−\mathrm{10}{t}−\mathrm{10}=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{10}{t}−\mathrm{10}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} +\mathrm{10}{t}−\mathrm{10}=\mathrm{0}\:\Rightarrow\mathrm{10}\left(\mathrm{1}−{t}\right)={t}^{\mathrm{2}} \\ $$$${x}−\sqrt{\mathrm{10}{x}}={x}\left(\mathrm{1}−\sqrt{\frac{\mathrm{10}}{{x}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{10}\left(\mathrm{1}−{t}\right)}{{t}^{\mathrm{2}} }=\mathrm{1}\:\checkmark \\ $$

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