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

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

Commented bybemath 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

Commented byAina Samuel Temidayo last updated on 19/Sep/20

This is not its only solution.

Commented bybemath last updated on 19/Sep/20

it only solution

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;+∞)

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,+∞)

Commented byruwedkabeh last updated on 19/Sep/20

Case II  −(x+1)

Commented byAina Samuel Temidayo last updated on 19/Sep/20

What was my error please?

Commented byAina Samuel Temidayo last updated on 19/Sep/20

Corrected. Thanks.^

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 ✓