Number of possible sollutions of x for the equation (2x+9)^2=x+3 ?

• (2x+9)^2=x+3
  =>4x^2+36x+81=x+3
  =>4x^2+35x+78=0
  now discriminant b^2-4ac=35^2-4*4*78=1225-1248=-23
  here b^2-4ac
  here Discriminant is negative...
  ans will be no real solution ony two imaginary solutions......
• 9 years agoHelpfull: Yes(7) No(2)
• (2x+9)^2=x+3
  =>4x^2+36x+81=x+3
  =>4x^2+35x+78=0
  now discriminant b^2-4ac=35^2-4*4*78=1225-1248=-23
  here b^2-4ac
• 9 years agoHelpfull: Yes(2) No(2)
• i think question is find the number of value of x where (2x+9)^2/(x+3) value is integer
  option a)2 b)4 c)6 4)infinity
• 9 years agoHelpfull: Yes(2) No(0)
• x=0,-2,-4,6,-6,-12 total
  
  
  x=6 possible solutions
• 9 years agoHelpfull: Yes(0) No(9)
• 0 Because If You Simplify This Equation You Will Get 4x^2 +35x+78 = 0. And By The Rule b^2-4ac = -23. So, Answer Is 0 Real Solutions.
• 9 years agoHelpfull: Yes(0) No(0)
• Since the equation is quadratic so the possible solution are 2 ,despite of either roots are real or imaginary.
• 9 years agoHelpfull: Yes(0) No(0)
• 2 real roots as the calculated delta comes out to be +ve
• 9 years agoHelpfull: Yes(0) No(1)
• Write (2x + 9 )^2 as (2x + 6 + 3) ^2 => ( 2(x +3) + 3 )^2 .
  Now Square it: 4 ( x+ 3) ^2 + 9 + 12(x+3) .
  Now divide this equation by x+3. ( as (2x+9)^2 = x+3 => (2x+9) / (x+3) =0 ) .
  So divide each term by x+3.
  Then take 9 / (x+3) [Do not consider other two terms as they can have infinite values for x] . So for this term to be real, x can have -12, and 6 as the value.
  So 2 values of X are possible.
• 8 years agoHelpfull: Yes(0) No(0)