* 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)