Consider the equation, 1−2x−5x2 The maximum value of above is given by, a=−
4+20
4(−5)
=
6
5
The minimum value is given by, b=−
4−20
4
= 4 Now the quadratic expression 5ax2+bx+7 at a=
6
5
and b=4 must be positive then, 6x2+4x+7>0 The discriminant is given by, D=16−4(6)(7) D=−152 The discriminant D<0 and coefficient of x2 is positive. So, 6x2+4x+7>0,∀x∈R=(−∞,∞)