root X+3=X root X+3 how many solution???

• root(x+3)=x root(x+3)
  =>square both side
  =>x+3=x2(x+3)
  =>(x+3)(x2-1)
  =>(x+3)(x+1)(x-1)
  hence 3 solution is there
• 8 years agoHelpfull: Yes(22) No(16)
• (X root x+3)- (root x+3)=0
  solutions are two
  1,-3
• 8 years agoHelpfull: Yes(17) No(5)
• it has three solutions (1,-1,-3)
  Solution:
  firstly squaring on both sides we get
  (X+3)=X^2(X+3)
  X^3+3(X^2)-X-3=0
  on simplifying this equation we get
  (X-1)((X^2)+4X+3)
  on solving this we get
  (X-1)(X+1)(X+3) Ans. 1,-1,-3 so three solution
  
• 8 years agoHelpfull: Yes(10) No(5)
• Given
  root(x+3)=x root(x+3)---------------(1)
  => Taking square both side
  =>x+3=x(x+3)^1/2
  =>(x+3)(x^2-1)
  =>(x+3)(x+1)(x-1)
  now check for each values values in the eq.1
  for x=-3 =>0=-3*0
  0=0 possible one sol.
  for x=-1 =>(-1+3)^1/2=-1*(-1+3)^1/2
  root 2=-root 2 not possible
  for x=-1 =>(1+3)^1/2=1*(1+3)^1/2
  2=2 possible 2nd sol.
  
  so total 2 solutions are possible
• 8 years agoHelpfull: Yes(9) No(3)
• 1 solution bec root x+3 cancel and only 1 value is possible
• 8 years agoHelpfull: Yes(7) No(6)
• only one solution
  since x=1 is satisfying above condition
  
• 8 years agoHelpfull: Yes(5) No(3)
• it has three solutions (1,-1,-3)
  Solution:
  firstly squaring on both sides we get
  (X+3)=X^2(X+3)
  X^3+3(X^2)-X-3=0
  on simplifying this equation we get
  (X-1)((X^2)+4X+3)
  on solving this we get
  (X-1)(X+1)(X+3) Ans. 1,-1,-3 so three solution
  
  
• 8 years agoHelpfull: Yes(3) No(4)
• there will two solution if we square both the side of equation.
  -1 & +1 so there is definitely 2 solution.
  here options are not given. so i can't say the correct ans
• 8 years agoHelpfull: Yes(2) No(4)
• root of(x+3)=x root of(x+3)
  square both side
  root(x+3)^2=[x root(x+3)]^2
  x+3=x^2*(x+3)
  (x+3)divided by (x+3)=x^2
  1=x^2
  x=+1,-1
  means two solution
  
  
• 8 years agoHelpfull: Yes(2) No(3)
• 2 solution.that is +1,-1
  
• 8 years agoHelpfull: Yes(1) No(0)
• When we are squaring both sides after finding out x we have to check whether LHS= RHS and if the condition satifies then the solution obtained is correct . In this case when we substitute X= -1 LHS# RHS and we have only 2 values for X
• 7 years agoHelpfull: Yes(1) No(0)
• the correct answer is 3
  explanation:
  root(x+3)-x root(x+3)=3
  root(x+3)[1-x]=3 take root(x+3) is common
  square both side
  we get,
  (x+3)(x-1)^2=9
  we get 3 value of x.
• 7 years agoHelpfull: Yes(1) No(0)
• 2 solution are possible
  rootx(x-1)=0
  x=1and x=0
  
• 8 years agoHelpfull: Yes(0) No(5)
• Only one solution possible - 1
• 8 years agoHelpfull: Yes(0) No(2)
• only one solution -3/2
  => x+3=x
  square both sides, we get
  => x^2+6x+9=x^2
  on simplifying it, we get
  => x= -3/2
• 8 years agoHelpfull: Yes(0) No(1)
• @Naveen Kumar we can not take G=8 as 8 is already given in the same row and we know every variable has unique value so it violates the basic rule of crypto.
  
  So look for the different values of G.
• 8 years agoHelpfull: Yes(0) No(2)
• Answer will be 0.
  Because other than 0 root will not be zero
• 8 years agoHelpfull: Yes(0) No(2)