If x^2-16>0,which of the following must be true nd how?
a)-4>x>4
b)-4>x

• for inequalities when you are taking square root of number remember following rules
  for x^2>16
  as x can be either positive or negative hence consider following cases
  1] if x is positive then x > 4 (no change of inequality)
  2] if x is negative then x < -4 (change of inequality)
  hence -4 > x > 4 is the correct answer but on TCS openseeme they given the wrong answer
• 12 years agoHelpfull: Yes(32) No(2)
• (a^2-b^2)=(a+b)(a-b)
  X^2-16=(x+4)(x-4)
  x+4=0 -->x=-4
  x-4=0 -->x=4
  so value of x lies between 4 and -4
  so a)is right ans
• 12 years agoHelpfull: Yes(13) No(16)
• the equation will be satisfy only if
  x belongs to (-infinity , -4)and (4 , infinity)
  values of option A is not satisfying equation.
  so the ans is option B.
• 12 years agoHelpfull: Yes(12) No(8)
• x2-16>0
  x2>16
  then x>-+4
  hence it is lies between -4>x>4
• 12 years agoHelpfull: Yes(7) No(4)
• x^2-4^2>0
  (x+4)(x-4)>0
  x+4>0,x-4>0
  
• 12 years agoHelpfull: Yes(4) No(5)
• x^2-16>0
  x^2-4^2>0
  (x-4)(x+4)>0
  x>4
  x>-4
  so the answer is a)-4>x>4
  
• 12 years agoHelpfull: Yes(2) No(11)
• its option [a] such a simple Q_ :)
• 12 years agoHelpfull: Yes(1) No(17)
• x^2-16>0
  (x-4)(x+4)>0
  then the value of x is selected such that X>4 and X0
  also -ve value of x say -5 then (-5)^2 -16=9>0
  so x>4 and x
• 12 years agoHelpfull: Yes(1) No(4)
• for x= -4 ans is not > than 0 but for value less than -4 it is > than 0
• 12 years agoHelpfull: Yes(1) No(5)
• The equation will satisfy for all numbers >4 and >-4. Hence the right answer will be X = >+/-4
• 11 years agoHelpfull: Yes(1) No(0)
• but is it ever possible that a number is less than -4 but greater than 4 , plz rectify me
  
• 10 years agoHelpfull: Yes(0) No(0)
• (x-4)(x+4)>0
  so,(x-4)>0 and (x+4)>0
  then
  x>4 and x
• 10 years agoHelpfull: Yes(0) No(0)
• the answer must be x4
• 10 years agoHelpfull: Yes(0) No(0)
• x^2-16>0
  x=+-4
  so x lies b/w -4and+4
  so ans is -4>x>+4
• 9 years agoHelpfull: Yes(0) No(0)