In this question, X^Y means, X raise to the power Y. How many integers x satisfy the equation (x^2-x-1)^(x+2)=1 ?

• the answer is 4 (0,-1,2,-2)
  if x=0
  (0-0-1)^(0+2)
  =(-1)^2
  =1
  
  if x=(-1)
  (1+1-1)^(2-1)
  =1^1
  =1
  
  if x=2
  (4-2-1)^(2+2)
  =1^4
  =1
  
  if x=(-2)
  we know that {(anything ^0)=1}
  we can put
  x+2=0
  or, x=-2
  
• 9 years agoHelpfull: Yes(51) No(1)
• I think x=0,-1,2,-2 will be answer so total 4 integers will satisfy the given equation.
• 9 years agoHelpfull: Yes(19) No(11)
• only three integer values are possible
  which are 0 , -1 , 2
  if we put 0 then
  (0-0-1)^(0+2) = 1
  if we put -1 then
  (1+1-1)^(-1+2) = 1
  if we put 2 then
  (4-2-1)^(2+2) = 1
  so hence only three integer values are possible.
• 9 years agoHelpfull: Yes(15) No(13)
• The answer will be..0,-1,2
• 9 years agoHelpfull: Yes(8) No(8)
• 0,-1,2,-2 totally 4 integers will satify first place 0 in place of x and then proceed like that
• 9 years agoHelpfull: Yes(6) No(7)
• Take log we get
  (x+2)log(x^2-x-1)=0 (since log1=0)
  x+2=0 or x^2-x-1=1 (if logx=0 then x=1)
  x=-2 or x(x-1)=2 => x=-1,2
  the ans will be x=-1,-2,2
  log of negative is undefined
• 9 years agoHelpfull: Yes(6) No(3)
• (x^2-x-1)^(x+2)=1
  This means either x+2=0 or x^2-x-1=1 for any (x+2) or x^2-x-1=-1 for even (x+2)
  x+2=0 -> x=-2
  x^2-x-1=1 -> x^2-x-2=0 -> (x-2)(x+1)=0 -> x=2,-1
  x^2-x-1=-1 -> x^2-x=0 -> x(x-1)=0 -> x=0,1. For x=0, x+2=2, i.e even. For x=1, x+2=3, i.e odd, hence discarded.
  Final Ans: x=-2,2,-1,0
• 9 years agoHelpfull: Yes(3) No(0)
• Answr is 1,-1,0,-2,2
  Fr answr to be 1 power shld be zero
• 9 years agoHelpfull: Yes(2) No(5)
• 0 -1 2-2 answer is 4
• 9 years agoHelpfull: Yes(2) No(0)
• 4 integers(0,-1,-2,2)
• 9 years agoHelpfull: Yes(2) No(0)
• explain please...
• 9 years agoHelpfull: Yes(1) No(1)
• please explain how did u find that
• 9 years agoHelpfull: Yes(1) No(0)
• As we know anything to the power 0 is 1
  Therefore, x+2=0 => x=-2
  Also, one to the power anything is 1
  Therefore, x^2 - x - 1 = 1
  x^2 - x - 2 = 0
  Solving the eq,
  We get
  x = 2, -1
  Also, zero satisfies the eq.
  Thus, 4 integers satisfy the eq (0,-1,2,-2)
• 9 years agoHelpfull: Yes(1) No(0)
• for satisfying the equation either x+2=0 or x^2-x-1=1 or x^2-x-1=-1 with an even power
  so value of x will be -2,2,-1,0
• 9 years agoHelpfull: Yes(0) No(1)
• At x = -1, => satisfied
  At x = 0, => satisfied
  At x = 1, => satisfied
  & for x > 1 & x < -1 => not satified the condition.
  so, answer is 3 integers..
• 9 years agoHelpfull: Yes(0) No(2)
• when we substitute 0,-1,-2 in the place of x then the given equation is satisfied
• 9 years agoHelpfull: Yes(0) No(0)
• Their are 4 integers ,0,2,-2,1
• 9 years agoHelpfull: Yes(0) No(0)
• 4 integers can satisfy this equation viz:0,2,-2,1
  make (x^2-x-1)=1 gives two values ie 2 n 1
  make x+1=1
  also put x=0 in eqn.
• 9 years agoHelpfull: Yes(0) No(0)
• I think answer will be 4. And values of x will be 0,-1,2,-2
• 9 years agoHelpfull: Yes(0) No(0)
• In the given equation substitute the value of x one by one with -2, -1, 0, 2
  We find these 4 integers satisfying the above condition.....Hence 4 is the answer...
• 9 years agoHelpfull: Yes(0) No(0)
• ans-2
  it is satisfied only incase of 0 nd 2.so ans 2
• 9 years agoHelpfull: Yes(0) No(1)
• Its ans is 4 for(-2,-1,0,2)
• 9 years agoHelpfull: Yes(0) No(0)
• since (x^2-x-1)^(x+2)=1,
  then we can write
  (x^2-x-1)^(x+2)=1^(x+2) (since 1 to the power any number is 1)
  so (x^2-x-1)^(x+2)=1
  solving we will get x=-1,2
  and solving the exponents we will get 0
  so total number of integer is 3
• 9 years agoHelpfull: Yes(0) No(0)
• (x^2-x-1)^(x+2)=1
  taking log both sides
  (x+2)log(x^2-x-1)=log1=0
  so 1 solution is x=-2
  x^2-x-1=e^0=1
  solving the equation
  x=2 x=-1
  so total 3 solutions
• 9 years agoHelpfull: Yes(0) No(0)