find the number of possible value of X.
for which (3*X+4)^2/(X+4)this gives the integer value.
options
a. 2 b.4 c. 6 d. infinite

• 6 values are possible.
  (3*X+4)^2/(X+4)
  = (2X+X+4)^2/(X+4)
  
  For integer value of (3*X+4)^2/(X+4)
  (2X)^2/(X+4) should be integer .
  For that (2X+8-8)^2)/(X+4) should be integer or 8^2/(X+4) should be integer.
  which is possible with 6 values of X like -2, 0,4,12,28,60.
  like X=0,4,12, 28,60
• 11 years agoHelpfull: Yes(10) No(15)
• @ Shresth,
  
  Pls check the
  
  u hv solved it as
  (3*X+4)^[2/(X+4)]
  I think,it should be solved as
  [(3*X+4)^2]/(X+4)
  
  If we consider
  (3*X+4)^[2/(X+4)], then x=0 also gives desired result as
  (3*0+4)^[2/(0+4)] = 4^(1/2)=2 which is integer.
  (3*X+4)^2/(X+4)
• 11 years agoHelpfull: Yes(6) No(1)
• 6 should be the ans. 3x+7 when divided by x+4 gives negative remainder 5.So,when (3x+7)^2 is divided,remainder net is 25.so,basically prob reduces to intg. values of x for which 25/X+4 gives 0 remainder.Clearly the values are (-29,-9,-5,-3,1,21).
• 11 years agoHelpfull: Yes(5) No(3)
• if x=4, then
  (3*4+4=16)
  and
  2/(4+4)=1/4
  now
  16^1/4 =4 which is integer.
  Except 4 , all are giving results in under roots which are not integer.
• 11 years agoHelpfull: Yes(3) No(2)
• wht is d rght ans?
  
• 11 years agoHelpfull: Yes(1) No(0)
• 14 values are possible x=-3,-5,60,-68,-2,-6,-8,0,4,-12,12,-20,28,-36
• 11 years agoHelpfull: Yes(1) No(0)
• ans infinite
• 8 years agoHelpfull: Yes(0) No(0)