In this question, X^Y means, X raised to the power Y. How many integers x satisfy the equation (x^2 - x - 1)^(x + 2)= 1?
o 4
o 2
o 3
o None of the other 3 choices

• answer is 4
  because from(x^2-x-1) you'll get 2
  values of x.
  and power should be 0 to get the 1
  x+2=0
  x=-2
  and if you put x=1
  then also you'll get 1 in rhs
  sono of total values of x is 4
• 8 years agoHelpfull: Yes(12) No(7)
• answer is 4 solutions ,x=-1,-2,0,2
• 8 years agoHelpfull: Yes(8) No(1)
• It should be 3 because only 0,-1 and 2 are satisfying this equation.
• 8 years agoHelpfull: Yes(3) No(4)
• 4 integer values will satisfy this equation
  for (x^2-x-1)^(x + 2)=1 there are certain cases :
  1. on equating both equations (x^2 - x - 1)^(x + 2)=1^1 then x+2=1 &x=-1
  2. value of (x^2-x-1) can be 1 or -1, (x^2-x-1)=1 equation will have real roots
  (x^2-x-1)=-1 will give x=0 & x=1, out of which x=0 will satisfy equation
  3. from 2nd case if (x^2-x-1)=-1 then x+2 can hold only even values i,e. 2,4,6,8....
  that will give x=0,2,4... but only x=0 & x=2 will satisfy the equation
  4. value of x+2 can be 0 i.e, x+2=0,x=-2
  so 4 integer values -2,-1,0,2 will satisfy the equation
• 8 years agoHelpfull: Yes(2) No(0)
• Taking log on both sides.
  >> (x+2).log(x^2-x-1)= log 1 =0.
  >>either (x+2) =0 or log(x^2-x-1)=0.
  x+2=0 >> x=-2.
  log(x^2-x-1)=0 >> (x^2-x-1)=e^0 = 1
  >> x^2-x-2=0
  >> x=5, -4.
  So the answer is 3, and the solutions are -2, -4, 5.
  
• 8 years agoHelpfull: Yes(1) No(2)
• 4
  ((-2)^2)-2-1)^(-2+2)=1
  ((0)^2)-0-1)^(0+2)=1
  ((2)^2)-2-1)^(2+2)=1
  ((-1)^2)-(-1)-1)^(-1+2)=1
• 8 years agoHelpfull: Yes(1) No(1)
• only -2,-1,0,2 satisfies hence answer is 4
  
• 8 years agoHelpfull: Yes(1) No(0)
• FIRST TAKE LOG BOTH THE SIDE IT WILL MAKE
  (x+2)log(x^2-x-1)=log1
  x+2=log1 log1= 0 x= -2
  then x^2-x-1=1
  x=,1,2
• 8 years agoHelpfull: Yes(0) No(0)
• (x^2-x-1)^(x+2)=1^0
  or,x+2=0
  or, x=-2
  x^2-x-1=1
  x=2,-1
  no of integers x satisfy=3
• 8 years agoHelpfull: Yes(0) No(1)