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# Test of Mathematics Solution Subjective 65 - Minimum Value of Quadratic  This is a Test of Mathematics Solution Subjective 65 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

## 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) This is a Test of Mathematics Solution Subjective 65 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

## 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)

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