how many solutions exists root(x+5)=xroot(x+5)

• root(x+5)= xroot(x+5)
  => root(x+5)- xroot(x+5) = 0
  => root(x+5)*(1-x)= 0
  => x = 1 , -5
  two soln.
• 9 years agoHelpfull: Yes(29) No(9)
• (x+5)=x^2(x+5)
  =(x+5)(x^2-1)=0
  =x^2-1=0
  =x^2=1
  =x=+1,-1
  x=-5
  thus 3 solution exists
• 9 years agoHelpfull: Yes(14) No(11)
• squaring both sides we get
  x^3+5x^2-x-5=0
  =>x^2(x+5)-1(x+5)=0
  =>(x+1)(x-1)(x+5)=0
  =>x=1,-1,-5
• 9 years agoHelpfull: Yes(5) No(2)
• And: 2. 1,-5 are only solutions. As -1 satisfies the equation after squaring but not the original one.
• 9 years agoHelpfull: Yes(4) No(0)
• x=1 only one solution exist
• 9 years agoHelpfull: Yes(3) No(4)
• x^2(x+5)=(x+5)
  (x+5)(x^2-1)=0
  x=-5
  x=+1
  x=-1
  hence there are three soln.
• 9 years agoHelpfull: Yes(3) No(1)
• root(x+5)=x root(x+5)
  we take square of both sides
  => (x+5)=x^2(x+5)--- (1)
  =>(x+5)(x^2-1)=0
  => x=-5, 1, -1
  but from eq (1), (x+5)/(x+5)=x^2; so, here x= -5 is not possible and when put x= -1 in original eq. then it doesn't satisfy.
  so, X= 1 (only 1 solution )
  
• 8 years agoHelpfull: Yes(3) No(0)
• puja sing ..question is correct
  
• 9 years agoHelpfull: Yes(1) No(0)
• options are 1 2 3 4
• 9 years agoHelpfull: Yes(0) No(1)
• (x+5)=x^2(x+5)
  x+5=x^3+5x^2
  so value of x can be upto 3 values
• 9 years agoHelpfull: Yes(0) No(0)
• @pooja bala singh,
  root(x+5) = x root(x+5)
  either both sides root(x+5) can be cancelled out an d x = 1
  or after squaring both sides (x+5) can be cancelled out and x^2 = 1 will gives two solns of x .i.e +-1.
  i am getting confused..
  
• 9 years agoHelpfull: Yes(0) No(0)
• @AK
  In this case we cant cancel anything bco'z we have to find maximum solution of x. So, We could also cancel term without squring.
• 9 years agoHelpfull: Yes(0) No(0)
• SOMETHING IS MISSING IN THIS QUESTION. WRITE IN A CORRECT WAY
  
• 9 years agoHelpfull: Yes(0) No(2)