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

• (x^2 - x - 1)^(x + 2) = 1
  at x = 0 => (-1)^2 = 1
  at x = -1 => (1 + 1 -1)^1 = 1
  at x = -2 => (4 + 2 - 1)^0 = 1
  at x = 2 => (4 - 2 - 1)^4 = 1
  hence 4 integers satisfy the equation i.e 0,-1,-2,2
  
  4 => Answer
• 9 years agoHelpfull: Yes(27) No(0)
• apply log on both side
  (x+2)log(x^2-x-1)=0
  either
  x+2=0;x=-2
  or
  x^2-x-1=1
  x=-1,2
  x=0
• 9 years agoHelpfull: Yes(5) No(0)
• 0 does not satisfy the equation as -ve values does not exist for logarithms
• 9 years agoHelpfull: Yes(2) No(0)
• 3 possible integers
  -2,2,-1
• 9 years agoHelpfull: Yes(2) No(0)
• 5
  x=0,1,-1,2,-2
  
• 9 years agoHelpfull: Yes(1) No(12)
• In My TCS OPEN SEE SAME TEST
  
  i kept 4 integers
  
  They evaluated it as correct
  
  So ans is 4 we want to consider 0,-1,-2,2
• 9 years agoHelpfull: Yes(1) No(0)
• where does x=0 comes from neha...pls xplain??
• 9 years agoHelpfull: Yes(0) No(0)
• 4
  x=-1,1,-2
  2-x-1=0
  x=-1
  x+2=0
  x=-2
  as 1^n=1
  so x=1
  
• 9 years agoHelpfull: Yes(0) No(1)
• 3
  solving we get x=2
  but test it for x=0 thus there will be 3 solutions
  
• 9 years agoHelpfull: Yes(0) No(0)
• what would be the ans 3 or 4 ?
• 9 years agoHelpfull: Yes(0) No(0)
• sub(0,-1,2) we will get the ans=1
• 9 years agoHelpfull: Yes(0) No(0)
• ans > o,-1,2,+2 so 4is the ans.
• 9 years agoHelpfull: Yes(0) No(0)