For how many integer x satisfy the equation :(x^2-x-1)^x+2 = 1?

• Ans is 4
  when x=0,-1,2,-2
  when x=-2 then
  power x+2= -2+2= 0
• 10 years agoHelpfull: Yes(16) No(12)
• only one integral soln , x=1
  
  (x^2-x-1)^x+2 = 1
  => (x^2-x-1)^x = -1
  
  => x = 1 only
• 10 years agoHelpfull: Yes(11) No(13)
• We have (x^2-x-1)^x+2 = 1
  so putting both side log we get
  (x+2)log(x^2-x-1)=log(1)
  => (x+2)log(x^2-x-1)=0
  =>either (x+2)=0 or log(x^2-x-1)=0 or both are 0
  solving both we get
  x+2=0 =>x=-2
  and (x^2-x-1)=1 [since log 1=0]
  => x= -1,2
  also x=0 satisfies the equation
  
  hence we have total 4 values
• 10 years agoHelpfull: Yes(8) No(0)
• ans is 3
  when x=0,-1,2
• 10 years agoHelpfull: Yes(7) No(10)
• With due respect i would like to request all people answering wrong to check their ans on paper and then post it to site !!
  it really generates a wasteful data and also waste time
• 10 years agoHelpfull: Yes(7) No(0)
• (x^2-x-1)^x=-1
  only 1 will satisfy this equation
  (1-1-1)^1=-1
  ans=1
• 10 years agoHelpfull: Yes(4) No(7)
• the ques is nt clr bcz its is nt clr that weather x+2 is in power or x is in power +2 seprate so both posssibilities gives diff reslts
• 10 years agoHelpfull: Yes(3) No(0)
• if the equation is a^b=1 then it will be true only if either a=1 or b=0 as anything raised to the power 0 equals to 1. hence the equation x^2-x-1 will be equal to 1 only for the values 0, -1 and 2. and x+2 will be equal to 0 for x=-2 hence 4 will be the true answer.... sumit kumar is right about it...
• 10 years agoHelpfull: Yes(3) No(0)
• ans is only 1.
  x=1 then (1^2-1-1)^1+2= (-1)^1+2=-1+2=1
  if you take x=0,-1,2,-2 the condition not satisfied
  when x=0=(0^2-0-1)^0+2=(-1)^0+2=1+2=3....... ........(-1)^0=+1 because anything power 0 is 1
  when x=-1=(-1^2-(-1)-1)^-1+2=(1+1-1)^-1+2=1^-1+2=1+2=3.............1^-1=1/(1^1)=1
  when x=2=(2^2-2-1)^2+2=(4-2-1)^2+2=1+2=3
  when x=-2=(-2^2-(-2)-1)^-2+2=(4+2-1)^-2+2=5^-2+2=1/(5^2)+2=0.04+2=2.04
  so only 1 is corect when x=1
  
• 10 years agoHelpfull: Yes(2) No(8)
• Ans is 3.
  When x=0,1,-1
• 10 years agoHelpfull: Yes(1) No(8)
• when x= 0,-1,2,-2
  ans is 4
• 10 years agoHelpfull: Yes(1) No(1)
• (x^2-x-1)^(x+2)=1
  
  The equation satisfies when x=0,-1,-2,2
  therefore,number of integers are 4
• 10 years agoHelpfull: Yes(1) No(0)
• May be 2 alone.
  4-2-1^4=1
• 10 years agoHelpfull: Yes(0) No(7)
• ans will be 4
• 10 years agoHelpfull: Yes(0) No(1)
• ans. is 4.
  for x=-2,-1,0,2
• 10 years agoHelpfull: Yes(0) No(2)
• but if we take x=2,then that value will b 1/64,then how the answer is 4??
• 10 years agoHelpfull: Yes(0) No(2)
• ans is 4.
  x+2=0
  x=-2,
  and x^2-x-1=1,
  then x=2,-1
  and x=0
• 10 years agoHelpfull: Yes(0) No(1)
• ans=3
  taking log on both side ans comes -2, 2, -1
• 10 years agoHelpfull: Yes(0) No(3)
• If the base is one the equation will be correct.
  Put x=2
  Then 1^4=1
• 10 years agoHelpfull: Yes(0) No(2)
• (x^2-x-1)^x+2=1
  (x^2-x-1)=1^(1/x+2)
  x^2-x-1=1
  x^2-x-2=0
  By solving it
  x=-1,2
  -1 satisfy the given equation so x=-1
• 10 years agoHelpfull: Yes(0) No(2)
• there is no such solution
  
• 10 years agoHelpfull: Yes(0) No(1)
• -1 is the answer.
  when x = -1 this equation gives the result as 1.
• 10 years agoHelpfull: Yes(0) No(1)
• 2 intergers are there.
  Explanation:
  (x^2-x-1)^x+2 = 1
  or,(x^2-x-1)^x = -1
  (-1)^any odd positive number=-1
  so x must not be any negative number or even number..and (x^2-x-1) should be -1.
  x^2-x-1=-1
  or, x^2-x=0
  or,x(x-1)=0
  so,x=0,x=1..
  So,2 integers are there to solve the equation.
  note:- (1)^negative number is not -1,it is +1.
• 10 years agoHelpfull: Yes(0) No(1)
• x=1 & x=-1 are the only solutions possible..therefore 2
• 10 years agoHelpfull: Yes(0) No(1)
• How the answer can be 4 or 3?? Ans will be only 1.
  If we put (x)^(-anything) it will be 1/(x) not -x. Then how can it be -1 or -2?????????
  Ans is 1..
• 10 years agoHelpfull: Yes(0) No(1)