How many integers x satisfy the equation

(x^2 - 6x + 10)x^2 + x = 1

• i think question is (x^2 - 6x + 10)x^2 + x = 1
  for any integer (x^2 - 6x + 10) and x^2 is always positive.
  also, x^2 is greater than x for any integer
  so (x2 - 6x + 10)x2 + x never equals to 1
  NO INTEGRAL SOLn
  ans 0
• 11 years agoHelpfull: Yes(21) No(1)
• I feel there are no integer solutions for this question.
  If it is (-2) instead of 2, there will be one integer solution that is x=3
• 11 years agoHelpfull: Yes(0) No(7)
• on solving the equation becomes
  2x^2 -11x+19 here d^2 = 121-4x19x2= negative hence no solution
• 11 years agoHelpfull: Yes(0) No(3)
• the left and right side must be equal for satisfiability if x=0 then both side is equal as the question has x^2+x as power to x^2-6x+10.if x=0 then (x*0-6*0+10)^0=1 as zero as power of any number is equals to one.which satisfy it.
• 11 years agoHelpfull: Yes(0) No(4)
• 0 as roots are imaginary
• 11 years agoHelpfull: Yes(0) No(0)
• no solution
  
• 11 years agoHelpfull: Yes(0) No(0)
• if we do trial and error with the integers from 1.
  the given equation becomes (1-6+10)1+1=1
  =>(5)2=1
  =>10=1
  which is not at all possible...
  so, i think no solution for this.
• 11 years agoHelpfull: Yes(0) No(0)