If $a, b, c$ are real & $a > 0$ , then the minimum value of $a x^{2} + b x + c$ , where $x$ is also real is

\frac{b^{2} - 4 a c}{4 a}

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\frac{4 a c - b^{2}}{4 a}

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\frac{b^{2} - 4 a c}{2 a}

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\frac{4 a c - b^{2}}{2 a}

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Solution

The correct option is A $\frac{4 a c - b^{2}}{4 a}$
Let $f (x) = a x^{2} + b x + c$ and given that $a, b, c$ are real and $a > 0$
$\Rightarrow f^{'} (x) = 2 a x + b = 0$
Least value is at $x = \frac{- b}{2 a}$
$\Rightarrow f (\frac{- b}{2 a}) = a {(\frac{- b}{2 a})}^{2} + b (\frac{- b}{2 a}) + c$
$= \frac{4 a c - b^{2}}{4 a}$
Hence, option B.

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Similar questions

Q. What are the roots of the quadratic equation

a x^{2} + b x + c = 0 ?

If b² – 4ac = 0 then The roots of the Quadratic equation ax² + bx + c = 0 are given by :

Q. Which of the given formulae is the correct one to find out the roots of the quadratic equation

a x^{2} + b x + c = 0

The roots of a quadratic equation $a x^{2} + b x + c = 0$ are given by $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , provided $b^{2} - 4 a c \geq 0$ .

Q. Prove that the roots of

a x^{2} + b x + c = 0

are

\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

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If a,b,c are real & a>0, then the minimum value of ax2+bx+c, where x is also real is

The correct option is A 4ac−b24aLet f(x)=ax2+bx+c and given that a,b,c are real and a>0⇒f′(x)=2ax+b=0Least value is at x=−b2a⇒f(−b2a)=a(−b2a)2+b(−b2a)+c =4ac−b24aHence, option B.

If $a, b, c$ are real & $a > 0$ , then the minimum value of $a x^{2} + b x + c$ , where $x$ is also real is