Q. How many solutions exist for (x+5)^1/3=x((x+5)^1/3)
Option
a) 1
b) 2
c) 3
d) 4

• (x+5)^1/3=x*[(x+5)^1/3]
  or (x+5)^1/3-x*[(x+5)^1/3]=0
  or (x+5)^1/3*(1-x)=0
  or (x+5)^1/3=0 ,1-x=0
  or x=-5 ,x=1
  only two solution
  option(2)
  
  
• 10 years agoHelpfull: Yes(44) No(1)
• option 2)-
  (x+5)^1/3-x(x+5)^1/3=0
  (x+5)^1/3[1-x]=0
  x=-5,1 means two solution
  
• 10 years agoHelpfull: Yes(6) No(0)
• Ans b) 2
  
  we can solve it in two ways:
  
  ONE:
  
  (x+5)^1/3=x((x+5)^1/3)
  
  (x+5)^1/3 - x((x+5)^1/3) = 0
  
  (x+5)^1/3(1-x) = 0
  
  (x+5)^1/3 = 0 or (1-x) = 0
  
  x = -5 or x = 1
  
  TWO:
  
  By Cubing both sides we get,
  
  (x+5)^1/3=x((x+5)^1/3)
  
  (x+5) = x^3 (x+5)
  
  By solving it, we get
  equation:
  x^4 + 5x^3 - x - 5 = 0
  
  x^3 (x + 5) - 1(x + 5) = 0
  
  (x^3 - 1)(x + 5) = 0
  
  x^3 - 1 =0 or x = -5
  
  x^3 = 1 => If it is square we can assume that, there would be two solutions +1 ,-1
  for odd powers, sign is same as before.
  
  
• 10 years agoHelpfull: Yes(5) No(1)
• rise to 3rd power
  (x+5)=x^3((x+5))
  degree=4
  so 4 solutions
• 10 years agoHelpfull: Yes(1) No(3)
• here (x+5)^1/3=x((x+5)1/3)
  x+5=x^3(x+5) {cubic on both side}
  x^3(x+5)-(x+5)=0
  (x+5)(x^3-1)=0
  now, x+5=0,x^3-1=0
  x=-5, (x-1)(x^2+x+1)=0
  x-1=0,x^2+x+1=0
  x=1,and x^2+x+1=0 have two roots
  x1=(-1+3^1/2i)/2
  x2=(-1-3^1/2)/2
  so total roots is -5,1,x1 and x2 ans=4
  
• 10 years agoHelpfull: Yes(0) No(0)
• 4
  sol:>
  cube both side we get
  (x+5)=x^3(x+5)
  =>(x+5)(x^3-1)=0
  x=-5,1,w,w^2 where
  w=(-1+(3^1/2)i)/2,w^2=(-1-(3^1/2)i)/2 i=(-1)^1/2
• 10 years agoHelpfull: Yes(0) No(0)
• this equality will be true only for x=1
  for x=-5
  it will cause (x+5)^1/3/(x+5)^1/3=0/0 form
  which is indefinite form so only x=1
• 10 years agoHelpfull: Yes(0) No(0)