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Question Number 204423 by maqsood last updated on 17/Feb/24
Commented by maqsood last updated on 17/Feb/24
plz solve
$$\:{plz}\:{solve} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Feb/24
x^2 +x(√(x+1)) −2(x+1)=0 let (√(x+1)) =y ⇒x=y^2 −1 (y^2 −1)^2 +y(y^2 −1)−2y^2 =0 y^4 −2y^2 +1+y^3 −y−2y^2 =0 y^4 +y^3 −4y^2 −y+1=0 y^2 +y−4−(1/y)+(1/y^2 )=0 (y−(1/y))^2 +(y−(1/y))−2=0 t^2 +t−2=0 (t+2)(t−1)=0 y−(1/y)=−2,1 y^2 +2y−1=0 ∨ y^2 −y−1=0 y=((−2±2(√2))/2),((1±(√5))/2) ∵ y=(√(x+1)) ≥0 ∴y=(√(x+1))=((−2+2(√2))/2),((1+(√5))/2) x+1=(((−2+2(√2))/2))^2 ,(((1+(√5))/2))^2 x=(−1+(√2) )^2 −1,(((1+(√5))/2))^2 −1 =1+2−2(√2) −1, ((1+5+2(√5) −4)/4) =2−2(√2) , ((2+2(√5))/4) {2−2(√2) ,((1+(√5))/2)}
$${x}^{\mathrm{2}} +{x}\sqrt{{x}+\mathrm{1}}\:−\mathrm{2}\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${let}\:\sqrt{{x}+\mathrm{1}}\:={y}\:\Rightarrow{x}={y}^{\mathrm{2}} −\mathrm{1} \\ $$$$\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +{y}\left({y}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{2}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}+{y}^{\mathrm{3}} −{y}−\mathrm{2}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}^{\mathrm{4}} +{y}^{\mathrm{3}} −\mathrm{4}{y}^{\mathrm{2}} −{y}+\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{4}−\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{1}}{{y}}\right)−\mathrm{2}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} +{t}−\mathrm{2}=\mathrm{0} \\ $$$$\left({t}+\mathrm{2}\right)\left({t}−\mathrm{1}\right)=\mathrm{0} \\ $$$${y}−\frac{\mathrm{1}}{{y}}=−\mathrm{2},\mathrm{1} \\ $$$${y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{1}=\mathrm{0}\:\vee\:{y}^{\mathrm{2}} −{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{−\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}},\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\because\:{y}=\sqrt{{x}+\mathrm{1}}\:\geqslant\mathrm{0} \\ $$$$\therefore{y}=\sqrt{{x}+\mathrm{1}}=\frac{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}},\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}+\mathrm{1}=\left(\frac{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} ,\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${x}=\left(−\mathrm{1}+\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} −\mathrm{1},\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\:\:\:=\mathrm{1}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\:−\mathrm{1},\:\frac{\mathrm{1}+\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{4}}{\mathrm{4}}\: \\ $$$$\:\:\:=\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\:,\:\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\left\{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\:,\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$
Commented by Frix last updated on 17/Feb/24
y=(√(x+1)) ⇒ y≥0 ⇒ only solutions fit the given equation
$${y}=\sqrt{{x}+\mathrm{1}}\:\Rightarrow\:{y}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{only}\:\:\mathrm{solutions}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$
Commented by Rasheed.Sindhi last updated on 17/Feb/24
Right sir, I′m going to edit my answer.Thanks!
$${Right}\:\boldsymbol{{sir}},\:{I}'{m}\:{going}\:{to}\:{edit}\:{my}\:{answer}.{Thanks}! \\ $$
Answered by Rasheed.Sindhi last updated on 17/Feb/24
x^2 +x(√(x+1)) −2(x+1)=0 Dividing by x^2 : 1+((√(x+1))/x)−2((((√(x+1)) )/x))^2 =0; x≠0 ((√(x+1))/x)=y 1+y−2y^2 =0 2y^2 −y−1=0 (y−1)(2y+1)=0 y=1,−1/2 ((√(x+1))/x)=1∨ ((√(x+1))/x)=−(1/2) { ((((√(x+1))/x)=1⇒x>0)),(( ((√(x+1))/x)=−(1/2)⇒x<0)) :} { ((x^2 =x+1; x>0)),(( x^2 =4(x+1); x<0)) :} { ((x^2 −x−1=0; x>0)),(( x^2 −4x−4=0; x<0)) :} { ((x=((1+(√5))/2))),((x=2−2(√2))) :}
$${x}^{\mathrm{2}} +{x}\sqrt{{x}+\mathrm{1}}\:−\mathrm{2}\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathcal{D}{ividing}\:{by}\:{x}^{\mathrm{2}} : \\ $$$$\mathrm{1}+\frac{\sqrt{{x}+\mathrm{1}}}{{x}}−\mathrm{2}\left(\frac{\sqrt{{x}+\mathrm{1}}\:}{{x}}\right)^{\mathrm{2}} =\mathrm{0};\:{x}\neq\mathrm{0} \\ $$$$\frac{\sqrt{{x}+\mathrm{1}}}{{x}}={y} \\ $$$$\mathrm{1}+{y}−\mathrm{2}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} −{y}−\mathrm{1}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left(\mathrm{2}{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$${y}=\mathrm{1},−\mathrm{1}/\mathrm{2} \\ $$$$\frac{\sqrt{{x}+\mathrm{1}}}{{x}}=\mathrm{1}\vee\:\frac{\sqrt{{x}+\mathrm{1}}}{{x}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\begin{cases}{\frac{\sqrt{{x}+\mathrm{1}}}{{x}}=\mathrm{1}\Rightarrow{x}>\mathrm{0}}\\{\:\frac{\sqrt{{x}+\mathrm{1}}}{{x}}=−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{x}<\mathrm{0}}\end{cases}\: \\ $$$$\begin{cases}{{x}^{\mathrm{2}} ={x}+\mathrm{1};\:{x}>\mathrm{0}}\\{\:{x}^{\mathrm{2}} =\mathrm{4}\left({x}+\mathrm{1}\right);\:{x}<\mathrm{0}}\end{cases}\:\: \\ $$$$\begin{cases}{{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0};\:{x}>\mathrm{0}}\\{\:{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4}=\mathrm{0};\:{x}<\mathrm{0}}\end{cases}\:\: \\ $$$$\begin{cases}{{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{{x}=\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}}\end{cases} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Feb/24
(x^2 /(x+1))+((x(√(x+1)) )/(x+1))−2=0 ((x/( (√(x+1)) )))^2 +(x/( (√(x+1))))−2=0 y^2 +y−2=0; y=(x/( (√(x+1)) )) (y+2)(y−1)=0 y=−2,1 { (((x/( (√(x+1)) ))=−2⇒x<0)),(((x/( (√(x+1)) ))=1⇒x>0)) :} { ((x^2 =(−2(√(x+1)) )^2 ;x<0)),((x^2 =((√(x+1)) )^2 ; x>0)) :} { ((x^2 −4x−4=0; x<0 )),((x^2 −x−1=0; x>0)) :} x_(<0) =((4−4(√2))/2)=2−2(√2) x_(>0) =((1+(√5))/2)
$$\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{1}}+\frac{{x}\sqrt{{x}+\mathrm{1}}\:}{{x}+\mathrm{1}}−\mathrm{2}=\mathrm{0} \\ $$$$\left(\frac{{x}}{\:\sqrt{{x}+\mathrm{1}}\:}\right)^{\mathrm{2}} +\frac{{x}}{\:\sqrt{{x}+\mathrm{1}}}−\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:{y}^{\mathrm{2}} +{y}−\mathrm{2}=\mathrm{0};\:{y}=\frac{{x}}{\:\sqrt{{x}+\mathrm{1}}\:} \\ $$$$\left({y}+\mathrm{2}\right)\left({y}−\mathrm{1}\right)=\mathrm{0} \\ $$$${y}=−\mathrm{2},\mathrm{1} \\ $$$$\begin{cases}{\frac{{x}}{\:\sqrt{{x}+\mathrm{1}}\:}=−\mathrm{2}\Rightarrow{x}<\mathrm{0}}\\{\frac{{x}}{\:\sqrt{{x}+\mathrm{1}}\:}=\mathrm{1}\Rightarrow{x}>\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} =\left(−\mathrm{2}\sqrt{{x}+\mathrm{1}}\:\right)^{\mathrm{2}} \:;{x}<\mathrm{0}}\\{{x}^{\mathrm{2}} =\left(\sqrt{{x}+\mathrm{1}}\:\right)^{\mathrm{2}} \:;\:{x}>\mathrm{0}}\end{cases}\: \\ $$$$\begin{cases}{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4}=\mathrm{0};\:{x}<\mathrm{0}\:}\\{{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0};\:{x}>\mathrm{0}}\end{cases}\: \\ $$$${x}_{<\mathrm{0}} =\frac{\mathrm{4}−\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$${x}_{>\mathrm{0}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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