find value 0f x
x^2+x-(a+2)(a+1)=0

• Answer is x=a+1;
  x^2+x=(a+2)(a+1)
  x(x+1)=(a+2)(a+1)
  x=a+1 ; x+1=a+2
  Eg:
  if a=1,then x=2,
  4+2-6=0
  0=0;
  
• 11 years agoHelpfull: Yes(4) No(2)
• x^2+x-(a+2)(a+1)=0
  => x^2+[(a+2)-(a+1)]*x-(a+2)(a+1)=0
  => x^2+(a+2)x-(a+1)x-(a+2)(a+1)=0
  => x[x+(a+2)]-(a+1)[x+(a+2)]=0
  => [x-(a+1)][x+(a+2)]=0
  => [x-(a+1)]=0 and [x+(a+2)]=0
  hence x=a+1 and x=-(a+2)
  and both the roots satisfy the given equation x^2+x-(a+2)(a+1)=0
  so there will be two roots of x namely (a+1) and -(a+2).
• 11 years agoHelpfull: Yes(3) No(1)
• x^2 + x - (a+2)(a+1) = 0
  if we have ax^2 + bx + c = 0
  then we can apply x = [-b +(or)- (b^2-4a*c)^1/2]/2a
  so x = [-1 +(or)- {(2a+1)(2a+5)}]1/2 /2
• 11 years agoHelpfull: Yes(0) No(0)
• Given x^2+x-(a+2)(a+1)=0
  >>>quadratic roots formula
  -b(+/-)sqrt(b^2-4ac)/2a
  
  >>>-1(+/-) sqrt(1-4(1)(-(a^2+3a+2)))/2
  >>>-1(+/-) sqrt(4a^2+12a+9)/2
  >>>-1(+/-) sqrt(2a+3)^2/2
  >>>-1(+/-) 2a+3/2...by solving
  
  >>>>x=a+1 0r x=-(a+2).
• 11 years agoHelpfull: Yes(0) No(0)