how many solutions exist for sqrt(x+5)=x(sqrt(x+5))
options are a)1 b)2 c)3 d)4

• x(sqrt(x+5))-sqrt(x+5)=0
  (x-1)(sqrt(x+5)=0
  so,
  x-1=0 or x+5=0;
  so,
  x=1 or -5
• 9 years agoHelpfull: Yes(31) No(3)
• ans: c i.e 3
  (x+5)^1/2 = x(x+5)^1/2
  squaring both side
  then we get
  x+5 = x^2(x+5)
  x+5 - x^2(x+5) = 0
  take x+5 as common
  x+5(1-x^2)=0
  (x+5)(1+x)(1-x)=0 i.e a^2-b^2
  x=-5;
  x=-1;
  x=1
• 9 years agoHelpfull: Yes(18) No(16)
• x=-5,-1,&1
  but
  x=-1 does not satisfy the eqn.
  so only 2 are ans
  (b)
• 9 years agoHelpfull: Yes(14) No(0)
• after calculation we get x^2=1 so, x=1 or -1. so, 2 solutions exist. ans: (b)
  
• 9 years agoHelpfull: Yes(3) No(3)
• ans is c ie 3
• 9 years agoHelpfull: Yes(2) No(7)
• Why -1 cannot be considered? if we put x=-1 in the equation we get:
  sqrt(-1+5)=-1(sqrt(-1+5))
  =>sqrt(4)=-1(sqrt(4))
  =>(+-)2=-1[(+-)2]
  =>(+-)2=(-+)2
  L.H.S.=R.H.S
  Why this caanot be considered?
• 9 years agoHelpfull: Yes(1) No(4)
• sqrt(x+5)=x sqrt(x+5)
  square both side,
  x+5=x^2 (x+5)
  x^3+5x^2+x+5=0
  (x^2+1)(x+5)=0
  x=1,-5
  x=1
• 9 years agoHelpfull: Yes(1) No(0)
• Come on guys!!! Ans is clearly 3. Let m explain in detail :
  By squaring on both sides you get : x+5 = x^2(x+5)
  ==> x+5 = x^3 + 5(x^2)
  
  ==> (x^3) + 5(x^2) - x - 5 = 0
  If you solve the above equation, we get 3 values for x i.e : +1, -1, -5 . Therefore, 3 solutions. Substitute them in the above equation, you will get the answer.
• 9 years agoHelpfull: Yes(1) No(2)
• Given sqrt(x+5)=x(sqrt(x+5)),
  Squaring both the sides, we get: (x+5)=x^2(x+5)
  Cancelling the equal terms i.e. (x+5) from both sides, we get: 1=x^2
  Now, taking the square root on both sides, we get
  1=x;
  which means either +1 or -1 can give 1,
  Thus it has 2 roots.
  
  Therefore , Answer is 2.
  Option B is the Answer.
• 5 years agoHelpfull: Yes(1) No(0)
• ans will be 3 soluton
  1st direcrtly cancel sqrt(x+5) frem both side we get x=1
  then bring above equestn in polynomial frm by squaring bth side
  then divide it by x-1 to get further roots of x
• 9 years agoHelpfull: Yes(0) No(2)
• ANSWER==> b)2
  x=1/-5
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 3
  suaring both side..x=5,-1,1
• 9 years agoHelpfull: Yes(0) No(3)
• c-3 solutions
• 9 years agoHelpfull: Yes(0) No(2)
• sq(x+5)=xsq(x+5) : squaring x+5 = x^2(x+5)
  x+5/(x+5)=X^2 , so x = +-1, so solution 2,,,
• 9 years agoHelpfull: Yes(0) No(1)
• we got x= -5,-1 and 1
  
  but put -1 in original eq:- sqrt(-1+5) =-1(sqrt(-1+5))
  -> 2 = -2
  so,-1 not satisfied the above equation so only two solution can be possible.
  Ans: b)2
• 9 years agoHelpfull: Yes(0) No(0)
• sqrt(x+5)=x(sqrt(x+5))
  squaring both sides
  (x+5)=x^2(x+5)
  x^2=1=>x=+/- 1
  so nly 2 solution
• 9 years agoHelpfull: Yes(0) No(0)
• sqrt(x+5)=x(sqrt(x+5))
  Squring both side
  x+5=x^3+5x
  => x^3+4x+-5=0
  =>(x-1)(x^2+x+5)=0
  x=1 is a solution
  and (x^2+x+5)=0
  D=b^2-4ac
  D=1-20=-19
  so there is no real solution apart from 1
  Correct answer is a) 1
• 5 years agoHelpfull: Yes(0) No(0)