[(6)times2t^(2)(l" beet "ax^(2)+bx+c=0,quad a!=0*1l" ones "],[2ti+t+314," di "c=.........]

If (4a+c)^(2)<=4b^(2) then one root of ax^(2)+bx+c=0 lies in

the quadratic equation 3ax^2 +2bx+c=0 has atleast one root between 0 and 1, if

If 2a+3b+6c =0 then the equation ax ^2 + bx +c=0 has atlest one root in

If 2a+3b+6c=0 , show that the equation ax^(2)+bx+c=0 has at least one root between 0 and 1.

If a+b + c = 0, then show that the quadratic equation 3ax^(2) + 2bx +c=0 has at least one root in [0,1].

If a,b,c are non-zero real numbers such that int_(0)^(1) ( 1+ cos^(8) x) (ax^(2) + bx + c ) dx = int _(0)^(2) (1+cos^(8)x) (ax^(2) + bx + c ) dx = 0 ,then the equation ax^(2) + bx +c = 0 will have

If 2a+3b+6c=0 then the equation ax^2+bx+c=0 has in the interval (0,1)

lim_(x rarr1)(ax^(2)+bx+c)/(cx^(2)+bx+a),a+b+c!=0