The solution set of ∣((x+1)/x)∣+∣x+1∣=(((x+1)^2 )/(∣x∣)) is


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Question Number 114422 by Aina Samuel Temidayo last updated on 19/Sep/20

$The solution set of$ $∣ \frac{x + 1}{x} ∣ + ∣ x + 1 ∣= \frac{{(x + 1)}^{2}}{∣ x ∣} is$
Commented by bemath last updated on 19/Sep/20
$((∣x+1∣)/(∣x∣))+∣x+1∣ = ((∣x+1∣^2 )/(∣x∣)) ; x≠0 ∣x+1∣ {∣x+1∣−∣x∣−1} = 0 { ((∣x+1∣=0→x=−1)),((∣x+1∣ = ∣x∣+1)) :} →x^2 +2x+1 = x^2 +2∣x∣+1 →2x = 2∣x∣ ; x = ∣x∣ →x>0 solution x=−1 ∪ x>0$
$\frac{∣ x + 1 ∣}{∣ x ∣} + ∣ x + 1 ∣ = \frac{∣ x + 1 ∣^{2}}{∣ x ∣}; x \neq 0$ $∣ x + 1 ∣ {∣ x + 1 ∣ - ∣ x ∣ - 1} = 0$ ${\begin{cases} ∣ x + 1 ∣= 0 \to x = - 1 \\ ∣ x + 1 ∣ = ∣ x ∣ + 1 \end{cases}$ $\to x^{2} + 2 x + 1 = x^{2} + 2 ∣ x ∣ + 1$ $\to 2 x = 2 ∣ x ∣; x = ∣ x ∣ \to x > 0$ $s o l u t i o n x = - 1 \cup x > 0$
Commented by Aina Samuel Temidayo last updated on 19/Sep/20

$This is not its only solution .$
Commented by bemath last updated on 19/Sep/20

$i t o n l y s o l u t i o n$
	Answered by 1549442205PVT last updated on 19/Sep/20
	$Solve the eqution ∣((x+1)/x)∣+∣x+1∣=(((x+1)^2 )/(∣x∣)) (1) We need the condition x≠0 We have the following tablet: determinant ((x,,(−1),,0,),((∣((x+1)/x)∣),((x+1)/x),0,(−((x+1)/x)),(∣∣),((x+1)/x)),((∣x+1∣),(−x−1),0,(x+1),1,(x+1)),((((x+1)^2 )/(∣x∣)),(−(((x+1)^2 )/x)),0,(−(((x+1)^2 )/x)),(∣∣),(((x+1)^2 )/x))) From above tablet we get i)If x≤−1 then (1)⇔((x+1)/x)−x−1=−(((x+1)^2 )/x) ⇔(x+1)^2 −x(x+1)+x+1=0 ⇔x^2 +2x+1−x^2 −x+x+1=0 ⇔2x+2=0⇔x+1=0⇔x=−1 ii)If −1<x<0 then (1)⇔−((x+1)/x)+x+1=−(((x+1)^2 )/x) ⇔(x+1)^2 +x(x+1)−(x+1)=0 ⇔x^2 +2x+1+x^2 +x−x−1=0 ⇔2x^2 +2x=0⇔2x(x+1)=0 ⇔x+1=0 (rejected as don′t satisfy ii)) iii)If x>0 then (1)⇔((x+1)/x)+x+1=(((x+1)^2 )/x) ⇔(x+1)^2 −x(x+1)−(x+1)=0 ⇔x^2 +2x+1−x^2 −x−x−1=0 ⇔0.x=0⇒∀x>0 are roots Combining three above cases we get the roots of given equation are x∈{−1}∪(0;+∞)$
	$Solve the eqution$ $∣ \frac{x + 1}{x} ∣ + ∣ x + 1 ∣= \frac{{(x + 1)}^{2}}{∣ x ∣} (1)$ $We need the condition x \neq 0$ $We have the following tablet :$ $\| \begin{matrix} x & - 1 & 0 \\ ∣ \frac{x + 1}{x} ∣ & \frac{x + 1}{x} & 0 & - \frac{x + 1}{x} & ∣∣ & \frac{x + 1}{x} \\ ∣ x + 1 ∣ & - x - 1 & 0 & x + 1 & 1 & x + 1 \\ \frac{{(x + 1)}^{2}}{∣ x ∣} & - \frac{{(x + 1)}^{2}}{x} & 0 & - \frac{{(x + 1)}^{2}}{x} & ∣∣ & \frac{{(x + 1)}^{2}}{x} \end{matrix} \|$ $From above tablet we get$ $i) If x ⩽ - 1 then$ $(1) \Leftrightarrow \frac{x + 1}{x} - x - 1 = - \frac{{(x + 1)}^{2}}{x}$ $\Leftrightarrow {(x + 1)}^{2} - x (x + 1) + x + 1 = 0$ $\Leftrightarrow x^{2} + 2 x + 1 - x^{2} - x + x + 1 = 0$ $\Leftrightarrow 2 x + 2 = 0 \Leftrightarrow x + 1 = 0 \Leftrightarrow x = - 1$ $ii) If - 1 < x < 0 then$ $(1) \Leftrightarrow - \frac{x + 1}{x} + x + 1 = - \frac{{(x + 1)}^{2}}{x}$ $\Leftrightarrow {(x + 1)}^{2} + x (x + 1) - (x + 1) = 0$ $\Leftrightarrow x^{2} + 2 x + 1 + x^{2} + x - x - 1 = 0$ $\Leftrightarrow {2 x}^{2} + 2 x = 0 \Leftrightarrow 2 x (x + 1) = 0$ $\Leftrightarrow x + 1 = 0 (rejected as {don}^{'} t satisfy ii))$ $iii) If x > 0 then$ $(1) \Leftrightarrow \frac{x + 1}{x} + x + 1 = \frac{{(x + 1)}^{2}}{x}$ $\Leftrightarrow {(x + 1)}^{2} - x (x + 1) - (x + 1) = 0$ $\Leftrightarrow x^{2} + 2 x + 1 - x^{2} - x - x - 1 = 0$ $\Leftrightarrow 0 . x = 0 \Rightarrow \forall x > 0 are roots$ $Combining three above cases we get$ $the roots of given equation are$ $x \in {- 1} \cup (0; + \infty)$
	Answered by Aina Samuel Temidayo last updated on 19/Sep/20
	$= ((∣x+1∣)/(∣x∣))+∣x+1∣ = (((x+1)^2 )/(∣x∣)), x≠0 CASE I: when x+1≥0 and x≥0 x≥−1 and x≥0 Finding their intersections ⇒ x≥0 but x≠0 ⇒ x>0 ⇒ ∣((x+1)/x)∣+∣x+1∣=(((x+1)^2 )/(∣x∣)) = ((x+1)/x)+x+1= ((x^2 +2x+1)/x) ⇒ x+1+x^2 +x= x^2 +2x+1 ⇒ 0=0 ⇒ x∈R Recall x>0 ⇒ x>0 CASE II: when x+1<0 and x<0 ⇒ x<−1 and x<0 ⇒ x<−1 ∣((x+1)/x)∣+∣x+1∣=(((x+1)^2 )/(∣x∣)) =((−(x+1))/(−(x)))+(−(x+1))= ((x^2 +2x+1)/(−(x))) ⇒−x−1+(−x(−(x+1)))=x^2 +2x+1 ⇒−x−1+(−x(−x−1))=x^2 +2x+1 ⇒−x−1+x^2 +x=x^2 +2x+1 ⇒x=−1 since x<−1 ⇒ x∈φ CASE III: when x+1<0 and x≥0 ⇒ x<−1 and x≥0 Finding their intersections ⇒ x∈φ CASE IV: when x+1≥0 and x<0 ⇒ x≥−1 and x<0 ⇒ x∈ [−1,0) ⇒∣((x+1)/x)∣+∣x+1∣=(((x+1)^2 )/(∣x∣)) ⇒((x+1)/(−x))+x+1=((x^2 +2x+1)/(−x)) ⇒(x+1)−x(x+1)=x^2 +2x+1 ⇒ x+1−x^2 −x=x^2 +2x+1 ⇒ 2x^2 +2x=0 ⇒x(x+1)=0 ⇒x=0 or x=−1 since x∈[−1,0) ⇒ x=−1 Combining Cases I,II,III and IV ⇒ x∈ (0,+∞) ∪ [−1] ⇒ x∈ {−1} ∪ (0,+∞)$
	$= \frac{∣ x + 1 ∣}{∣ x ∣} + ∣ x + 1 ∣ = \frac{{(x + 1)}^{2}}{∣ x ∣}, x \neq 0$ $CASE I :$ $when x + 1 ⩾ 0 and x ⩾ 0$ $x ⩾ - 1 and x ⩾ 0$ $Finding their intersections$ $\Rightarrow x ⩾ 0$ $but x \neq 0$ $\Rightarrow x > 0$ $\Rightarrow$ $∣ \frac{x + 1}{x} ∣ + ∣ x + 1 ∣= \frac{{(x + 1)}^{2}}{∣ x ∣}$ $= \frac{x + 1}{x} + x + 1 = \frac{x^{2} + 2 x + 1}{x}$ $\Rightarrow x + 1 + x^{2} + x = x^{2} + 2 x + 1$ $\Rightarrow 0 = 0$ $\Rightarrow x \in R$ $Recall x > 0$ $\Rightarrow x > 0$ $CASE II :$ $when x + 1 < 0 and x < 0$ $\Rightarrow x < - 1 and x < 0$ $\Rightarrow x < - 1$ $∣ \frac{x + 1}{x} ∣ + ∣ x + 1 ∣= \frac{{(x + 1)}^{2}}{∣ x ∣}$ $= \frac{- (x + 1)}{- (x)} + (- (x + 1)) = \frac{x^{2} + 2 x + 1}{- (x)}$ $\Rightarrow - x - 1 + (- x (- (x + 1))) = x^{2} + 2 x + 1$ $\Rightarrow - x - 1 + (- x (- x - 1)) = x^{2} + 2 x + 1$ $\Rightarrow - x - 1 + x^{2} + x = x^{2} + 2 x + 1$ $\Rightarrow x = - 1$ $since x < - 1$ $\Rightarrow x \in ϕ$ $CASE III :$ $when x + 1 < 0 and x ⩾ 0$ $\Rightarrow x < - 1 and x ⩾ 0$ $Finding their intersections$ $\Rightarrow x \in ϕ$ $CASE IV :$ $when x + 1 ⩾ 0 and x < 0$ $\Rightarrow x ⩾ - 1 and x < 0$ $\Rightarrow x \in [- 1, 0)$ $\Rightarrow∣ \frac{x + 1}{x} ∣ + ∣ x + 1 ∣= \frac{{(x + 1)}^{2}}{∣ x ∣}$ $\Rightarrow \frac{x + 1}{- x} + x + 1 = \frac{x^{2} + 2 x + 1}{- x}$ $\Rightarrow (x + 1) - x (x + 1) = x^{2} + 2 x + 1$ $\Rightarrow x + 1 - x^{2} - x = x^{2} + 2 x + 1$ $\Rightarrow {2 x}^{2} + 2 x = 0$ $\Rightarrow x (x + 1) = 0$ $\Rightarrow x = 0 or x = - 1$ $since x \in [- 1, 0)$ $\Rightarrow x = - 1$ $Combining Cases I, II, III and IV$ $\Rightarrow x \in (0, + \infty) \cup [- 1]$ $\Rightarrow x \in {- 1} \cup (0, + \infty)$
	Commented by ruwedkabeh last updated on 19/Sep/20

	$C a s e I I$ $- (x + 1)$
	Commented by Aina Samuel Temidayo last updated on 19/Sep/20

	$What was my error please ?$
	Commented by Aina Samuel Temidayo last updated on 19/Sep/20

	$Corrected . Thanks \overset{}{.}$
	Answered by mnjuly1970 last updated on 19/Sep/20
	$solution::(x+1)^2 =∣x+1∣^2 ∣x+1∣((1/(∣x∣))+1−((∣x+1∣)/(∣x∣)))=0 x=−1 ✓ or (1/(∣x∣))+1−((∣x+1∣)/(∣x∣))=0 if x<−1 ⇒−(1/x)+1+[((x+1)/x)=1+(1/x)]=0 2=0 impossible if −1≤x≤0 ⇒ −(1/x)+1−1−(1/x)=0⇒((−2)/x)=0 again it is impossible. x>0⇒ (1/x)+1−1−(1/x)=0 ✓✓ ans {−1}∪ x>0 ✓$
	$s o l u t i o n :: {(x + 1)}^{2} =∣ x + 1 ∣^{2}$ $∣ x + 1 ∣ (\frac{1}{∣ x ∣} + 1 - \frac{∣ x + 1 ∣}{∣ x ∣}) = 0$ $x = - 1 ✓ o r \frac{1}{∣ x ∣} + 1 - \frac{∣ x + 1 ∣}{∣ x ∣} = 0$ $i f x < - 1 \Rightarrow - \frac{1}{x} + 1 + [\frac{x + 1}{x} = 1 + \frac{1}{x}] = 0$ $2 = 0 i m p o s s i b l e$ $i f - 1 ⩽ x ⩽ 0 \Rightarrow - \frac{1}{x} + 1 - 1 - \frac{1}{x} = 0 \Rightarrow \frac{- 2}{x} = 0$ $a g a i n i t i s i m p o s s i b l e .$ $x > 0 \Rightarrow \frac{1}{x} + 1 - 1 - \frac{1}{x} = 0 ✓ ✓$ $a n s {- 1} \cup x > 0 ✓$
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