How many integers satisfy the equation
(x^2-x-1)^(x+2)=1
a)none
b)3
c)2
d)4

• D is the answer.
  Try and error question.
  Equ satisfy by x=-2,2,0,-1
• 10 years agoHelpfull: Yes(28) No(2)
• d is correct (1,-1,2,0)
• 10 years agoHelpfull: Yes(6) No(8)
• take log both side
  (x+2)log(x^2-x-1)=log1=0
  case1
  x+2=0
  x=-2
  or
  case 2
  log(x^2-x-1)=0
  taking antilog
  x^2-x-1=10^0
  x^2-x-1=1
  x^2-x-2=0
  x=2,-1;
  hence ans is 3;
• 10 years agoHelpfull: Yes(3) No(1)
• ans-4
  
  -2,-1,0,2
• 10 years agoHelpfull: Yes(2) No(2)
• ans b)
  the values for x are -1, -2, 2
  0 is incorrect, because putting x=0 in (x^2 - x - 1) will make it -ve and the range of (x^2 - x - 1) should be positive (here).
• 10 years agoHelpfull: Yes(2) No(9)
• x may be -1, 2, -2 or 0
  if x=-1: ((-1)^2 +1 -1)^(-1+2)= 1^1 =1
  if x=2: ((2)^2+2-1)^(-2+2)=5^0=1
  etc.....
  Therefore 4 integers are possible.
  ans:d
• 10 years agoHelpfull: Yes(2) No(0)
• ans:c
  (2^2-2-1)^(2+2)
  (4-3)^(4)
  1^4=1
• 10 years agoHelpfull: Yes(2) No(1)
• put x=0 we get 1=1 so x=1 is correct and also
  applying log on both side
  (x+2)log(x^2-x-1)=log(1)=0
  i.e.,x=-2 or
  log(x^2-x-1)=0 => x^2-x-1=1 =>x^2-x=2 => x(x-1)=2 => x=2,-1.
  therefore ans is 4 integers (-2,-1,0,2)
• 9 years agoHelpfull: Yes(2) No(0)
• now there are five integers -2 -1 0 1 2
  so none (a).
• 10 years agoHelpfull: Yes(1) No(5)
• 3 cases arises.
  1.x^2-x-1=1 ============>-1,2
  2.x+2=0 ===============>-2
  3.x^2-x-1=-1 and x+2=2 => 0
  hence 4 solutions.
• 10 years agoHelpfull: Yes(1) No(0)
• 3 integral values
  
• 10 years agoHelpfull: Yes(1) No(0)
• a. 3.--》 -1, 0, 2.
• 10 years agoHelpfull: Yes(1) No(0)
• ans:d)4
  -1,0,2,-2
• 10 years agoHelpfull: Yes(0) No(0)
• ans is d
  (-1,-2,0,2)
• 10 years agoHelpfull: Yes(0) No(0)
• ans:c (0,2)
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• D is the corect answer for this
• 10 years agoHelpfull: Yes(0) No(0)
• (x^2-x-1) must be equal to 1
  by solving we get 2 and -1 as roots so ans is 2
• 10 years agoHelpfull: Yes(0) No(0)
• answer is d)4
  x=0,2,-2,-1
• 10 years agoHelpfull: Yes(0) No(2)
• ans 2,,,,,(x^2-x-1)^(x+2)=1 base should be 1 or -1 take x^2-x-1 =1 or -1,,then x^2-x=0, x=0 ,x=1
• 10 years agoHelpfull: Yes(0) No(0)
• the ans is :none
  as the number 1 alone satisfies the equation remaing integers fail to satisfy. so ans is none
  
• 10 years agoHelpfull: Yes(0) No(1)
• answer is 4
• 10 years agoHelpfull: Yes(0) No(0)
• ans :- c
  (2^2-2-1)^(2+2) =1
  (4-2-1)^4 =1
  (4-3)^4 =1
  (1)^4 =1
  
• 10 years agoHelpfull: Yes(0) No(0)
• ans b)
  the values for which x satisfies are -1,2,2.
  Althogh it satisfies 0,but it is not a solution because apply log on both sides
  (x+2)log(x^2-x-1)=0
  here when x=0,log (-1) which is not defined.
• 9 years agoHelpfull: Yes(0) No(0)
• C is the answer
  (x^2-x-1)^(x+2)=1
  By solving this we get x=-1 and x=3
  son the range is -1 to 3 but -1 and 3 only results in value = 1 , others values like 0, 1 and 2 results in fractional values -1/2, -1/3 and -4. Thus the answer is c) 2.
• 9 years agoHelpfull: Yes(0) No(0)
• C x=-1,-2
• 9 years agoHelpfull: Yes(0) No(0)
• check it by integer
  
• 9 years agoHelpfull: Yes(0) No(0)
• 3 cases gives solutions.
  1. x^2-x-1=1 =>-1,2
  2. x+2=0 =>-2
  3. x^2-x-1=-1 and x+2=2 => 0
  hence 4 solutions.
• 9 years agoHelpfull: Yes(0) No(0)
• we need to satisfy (x^2-x-1)^(x+2)=1 (x^2-x-1) must b 1
  satisfied values are::x=2,x=0
  -> 2 values
  ans:c
  
• 8 years agoHelpfull: Yes(0) No(0)