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Question Number 204423 by maqsood last updated on 17/Feb/24

Commented by maqsood last updated on 17/Feb/24

$p l z s o l v e$
	Answered by Rasheed.Sindhi last updated on 17/Feb/24
	$x^2 +x(√(x+1)) −2(x+1)=0 let (√(x+1)) =y ⇒x=y^2 −1 (y^2 −1)^2 +y(y^2 −1)−2y^2 =0 y^4 −2y^2 +1+y^3 −y−2y^2 =0 y^4 +y^3 −4y^2 −y+1=0 y^2 +y−4−(1/y)+(1/y^2 )=0 (y−(1/y))^2 +(y−(1/y))−2=0 t^2 +t−2=0 (t+2)(t−1)=0 y−(1/y)=−2,1 y^2 +2y−1=0 ∨ y^2 −y−1=0 y=((−2±2(√2))/2),((1±(√5))/2) ∵ y=(√(x+1)) ≥0 ∴y=(√(x+1))=((−2+2(√2))/2),((1+(√5))/2) x+1=(((−2+2(√2))/2))^2 ,(((1+(√5))/2))^2 x=(−1+(√2) )^2 −1,(((1+(√5))/2))^2 −1 =1+2−2(√2) −1, ((1+5+2(√5) −4)/4) =2−2(√2) , ((2+2(√5))/4) {2−2(√2) ,((1+(√5))/2)}$
	$x^{2} + x \sqrt{x + 1} - 2 (x + 1) = 0$ $l e t \sqrt{x + 1} = y \Rightarrow x = y^{2} - 1$ ${(y^{2} - 1)}^{2} + y (y^{2} - 1) - 2 y^{2} = 0$ $y^{4} - 2 y^{2} + 1 + y^{3} - y - 2 y^{2} = 0$ $y^{4} + y^{3} - 4 y^{2} - y + 1 = 0$ $y^{2} + y - 4 - \frac{1}{y} + \frac{1}{y^{2}} = 0$ ${(y - \frac{1}{y})}^{2} + (y - \frac{1}{y}) - 2 = 0$ $t^{2} + t - 2 = 0$ $(t + 2) (t - 1) = 0$ $y - \frac{1}{y} = - 2, 1$ $y^{2} + 2 y - 1 = 0 \lor y^{2} - y - 1 = 0$ $y = \frac{- 2 \pm 2 \sqrt{2}}{2}, \frac{1 \pm \sqrt{5}}{2}$ $∵ y = \sqrt{x + 1} ⩾ 0$ $∴ y = \sqrt{x + 1} = \frac{- 2 + 2 \sqrt{2}}{2}, \frac{1 + \sqrt{5}}{2}$ $x + 1 = {(\frac{- 2 + 2 \sqrt{2}}{2})}^{2}, {(\frac{1 + \sqrt{5}}{2})}^{2}$ $x = {(- 1 + \sqrt{2})}^{2} - 1, {(\frac{1 + \sqrt{5}}{2})}^{2} - 1$ $= 1 + 2 - 2 \sqrt{2} - 1, \frac{1 + 5 + 2 \sqrt{5} - 4}{4}$ $= 2 - 2 \sqrt{2}, \frac{2 + 2 \sqrt{5}}{4}$ ${2 - 2 \sqrt{2}, \frac{1 + \sqrt{5}}{2}}$
	Commented by Frix last updated on 17/Feb/24

	$y = \sqrt{x + 1} \Rightarrow y ⩾ 0$ $\Rightarrow only solutions fit the given equation$
	Commented by Rasheed.Sindhi last updated on 17/Feb/24

	$R i g h t s i r, I^{'} m g o i n g t o e d i t m y a n s w e r . T h a n k s!$
	Answered by Rasheed.Sindhi last updated on 17/Feb/24
	$x^2 +x(√(x+1)) −2(x+1)=0 Dividing by x^2 : 1+((√(x+1))/x)−2((((√(x+1)) )/x))^2 =0; x≠0 ((√(x+1))/x)=y 1+y−2y^2 =0 2y^2 −y−1=0 (y−1)(2y+1)=0 y=1,−1/2 ((√(x+1))/x)=1∨ ((√(x+1))/x)=−(1/2) { ((((√(x+1))/x)=1⇒x>0)),(( ((√(x+1))/x)=−(1/2)⇒x<0)) :} { ((x^2 =x+1; x>0)),(( x^2 =4(x+1); x<0)) :} { ((x^2 −x−1=0; x>0)),(( x^2 −4x−4=0; x<0)) :} { ((x=((1+(√5))/2))),((x=2−2(√2))) :}$
	$x^{2} + x \sqrt{x + 1} - 2 (x + 1) = 0$ $D i v i d i n g b y x^{2} :$ $1 + \frac{\sqrt{x + 1}}{x} - 2 {(\frac{\sqrt{x + 1}}{x})}^{2} = 0; x \neq 0$ $\frac{\sqrt{x + 1}}{x} = y$ $1 + y - 2 y^{2} = 0$ $2 y^{2} - y - 1 = 0$ $(y - 1) (2 y + 1) = 0$ $y = 1, - 1 / 2$ $\frac{\sqrt{x + 1}}{x} = 1 \lor \frac{\sqrt{x + 1}}{x} = - \frac{1}{2}$ ${\begin{cases} \frac{\sqrt{x + 1}}{x} = 1 \Rightarrow x > 0 \\ \frac{\sqrt{x + 1}}{x} = - \frac{1}{2} \Rightarrow x < 0 \end{cases}$ ${\begin{cases} x^{2} = x + 1; x > 0 \\ x^{2} = 4 (x + 1); x < 0 \end{cases}$ ${\begin{cases} x^{2} - x - 1 = 0; x > 0 \\ x^{2} - 4 x - 4 = 0; x < 0 \end{cases}$ ${\begin{cases} x = \frac{1 + \sqrt{5}}{2} \\ x = 2 - 2 \sqrt{2} \end{cases}$
	Answered by Rasheed.Sindhi last updated on 18/Feb/24
	$(x^2 /(x+1))+((x(√(x+1)) )/(x+1))−2=0 ((x/( (√(x+1)) )))^2 +(x/( (√(x+1))))−2=0 y^2 +y−2=0; y=(x/( (√(x+1)) )) (y+2)(y−1)=0 y=−2,1 { (((x/( (√(x+1)) ))=−2⇒x<0)),(((x/( (√(x+1)) ))=1⇒x>0)) :} { ((x^2 =(−2(√(x+1)) )^2 ;x<0)),((x^2 =((√(x+1)) )^2 ; x>0)) :} { ((x^2 −4x−4=0; x<0 )),((x^2 −x−1=0; x>0)) :} x_(<0) =((4−4(√2))/2)=2−2(√2) x_(>0) =((1+(√5))/2)$
	$\frac{x^{2}}{x + 1} + \frac{x \sqrt{x + 1}}{x + 1} - 2 = 0$ ${(\frac{x}{\sqrt{x + 1}})}^{2} + \frac{x}{\sqrt{x + 1}} - 2 = 0$ $y^{2} + y - 2 = 0; y = \frac{x}{\sqrt{x + 1}}$ $(y + 2) (y - 1) = 0$ $y = - 2, 1$ ${\begin{cases} \frac{x}{\sqrt{x + 1}} = - 2 \Rightarrow x < 0 \\ \frac{x}{\sqrt{x + 1}} = 1 \Rightarrow x > 0 \end{cases}$ ${\begin{cases} x^{2} = {(- 2 \sqrt{x + 1})}^{2}; x < 0 \\ x^{2} = {(\sqrt{x + 1})}^{2}; x > 0 \end{cases}$ ${\begin{cases} x^{2} - 4 x - 4 = 0; x < 0 \\ x^{2} - x - 1 = 0; x > 0 \end{cases}$ $x_{< 0} = \frac{4 - 4 \sqrt{2}}{2} = 2 - 2 \sqrt{2}$ $x_{> 0} = \frac{1 + \sqrt{5}}{2}$
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