problem: Show that for all real x, the expression {ax^2} + bx + C ( where a, b, c are real constants with a > 0), has the minimum value {\frac{4ac - b^2}{4a}} . Also find the value of x for which this minimum value is attained.

solution: f (x) {ax^2} + bx + c

Now minimum derivative = 0 & 2nd order derivative > 0.

{\frac{df(x)}{dx}} = 2ax + b
Or {\frac{d^2f(x)}{dx^2}} =2a
Now 2a> so 2nd order derivative > 0 so {\frac{d^2f(x)}{dx^2}} = 2.

So minimum occurs when

{\frac{df(x)}{d(x)}} = 0 or 2ax + b = 0

or 2ax = -b
or x = {\frac{-b}{2a}} (ans)

At x = {\frac{-b}{2a}}

{ax^2} + bx + c
= {a\times {\frac{b^4}{4a^2}}} + {b\times {\frac{-b}{2a}}} + c
= {\frac{4ac-b^2}{4a}} (proved)